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The asymptotic expansion of an integral of an exponential function
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What is the asymptotic expansion of $f(x) := int_0^1 e^-x(1-u^2)du$?
The integrand steeply declines near $u=1$. I tried to transform $u$ into something that is suitable for the method of steepest descent, but have not found an appropriate transformation. Integration by parts has not yield a satisfactory result, perhaps due to I having not found the right components to integrate.
real-analysis asymptotics
asked Aug 22 at 4:21
Hans
4,18221028
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up vote
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What is the asymptotic expansion of $f(x) := int_0^1 e^-x(1-u^2)du$?
The integrand steeply declines near $u=1$. I tried to transform $u$ into something that is suitable for the method of steepest descent, but have not found an appropriate transformation. Integration by parts has not yield a satisfactory result, perhaps due to I having not found the right components to integrate.
real-analysis asymptotics
asked Aug 22 at 4:21
Hans
4,18221028
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
What is the asymptotic expansion of $f(x) := int_0^1 e^-x(1-u^2)du$?
The integrand steeply declines near $u=1$. I tried to transform $u$ into something that is suitable for the method of steepest descent, but have not found an appropriate transformation. Integration by parts has not yield a satisfactory result, perhaps due to I having not found the right components to integrate.
real-analysis asymptotics
asked Aug 22 at 4:21
Hans
4,18221028
What is the asymptotic expansion of $f(x) := int_0^1 e^-x(1-u^2)du$?
The integrand steeply declines near $u=1$. I tried to transform $u$ into something that is suitable for the method of steepest descent, but have not found an appropriate transformation. Integration by parts has not yield a satisfactory result, perhaps due to I having not found the right components to integrate.
real-analysis asymptotics
asked Aug 22 at 4:21
Hans
4,18221028
edited Aug 22 at 8:34
asked Aug 22 at 4:21
Hans
4,18221028
asked Aug 22 at 4:21
Hans
4,18221028
asked Aug 22 at 4:21
Hans
4,18221028
4,18221028
add a comment |Â
add a comment |Â
2 Answers
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up vote
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The minimum of $1-u^2$ occurs at $u=1$, and near there we have $1-u^2 approx 2(1-u)$, which is linear in the quantity $1-u$. This suggests we make the change of variables $1-u^2 = v$ (which is linear in $v$), giving
$$
int_0^1 e^-x(1-u^2),du = frac12 int_0^1 e^-xv (1-v)^-1/2,dv.
$$
Then, following Watson's lemma, we can get the asymptotic expansion by expanding the subdominant term, $(1-v)^-1/2$, around $v=0$ and integrating term-by-term from $v=0$ to $v=infty$:
$$
beginalign
int_0^1 e^-x(1-u^2),du &= frac12 int_0^1 e^-xv (1-v)^-1/2,dv \
&approx frac12 sum_k=0^infty binom-1/2k (-1)^k int_0^infty e^-xv v^k,dv \
&= frac12 sum_k=0^infty binom-1/2k frac(-1)^k k!x^k+1
endalign
$$
as $x to infty$, which matches the series given in the other answer.
Let's prove the special case of Watson's lemma we use in this answer. We'll assume that $g(v)$ is analytic at $v=0$ and that $int_0^a lvert g(v) rvert,dv$ exists (these are certainly true for $g(v) = (1-v)^-1/2$), and show that
$$
I(v) = int_0^a e^-xvg(v),dv approx sum_k=0^infty fracg^(k)(0)x^k+1
$$
as $x to infty$. In other words, we will show that the asymptotic expansion for the integral can be obtained by expanding $g(v)$ in Taylor series around $v=0$ and integrating term-by-term.
Since $g(v)$ is analytic at $v=0$, there is a $delta in (0,a]$ such that $g^(k)(v)$ is analytic on $[0,delta]$ for all $k$. Further, Taylor's theorem tells us that for any positive integer $N$ and any $v in [0,delta]$, there is a $v^* in [0,v]$ for which
$$
g(v) = sum_k=0^N fracg^(k)(0)k! v^k + fracg^(N+1)(v^*)(N+1)! v^N+1. tag1
$$
We will split the integral $I(v)$ at this $delta$, and estimate each piece separately. To this end, we define
$$
I(v) = int_0^delta e^-xvg(v),dv + int_delta^a e^-xvg(v),dv = I_1(v) + I_2(v).
$$
We only need a rough estimate on $I_2(v)$:
$$
lvert I_2(v) rvert leq int_delta^a e^-xv lvert g(v) rvert ,dv leq e^-delta x int_0^a lvert g(v) rvert,dv,
$$
where we have assumed the last integral on the right is finite. Thus
$$
I(v) = I_1(v) + O(e^-delta x)
$$
as $x to infty$.
Now, from $(1)$ we have
$$
I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + frac1(N+1)! int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv.
$$
The last integral can be bounded by
$$
leftlvert int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv rightrvert leq left( sup_0 < v < delta leftlvert g^(N+1)(v) rightrvert right) int_0^infty e^-xv v^N+1,dv = fractextconst.x^N+2.
$$
Thus
$$
I_1(v) = I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + O!left(x^-N-2right)
$$
as $x to infty$. Finally we reattach the tails to the integrals in the sum,
$$
beginalign
int_0^delta e^-xv v^k ,dv &= int_0^infty e^-xv v^k,dv - int_delta^infty e^-xv v^k,dv \
&= frack!x^k+1 + O!left(e^-delta xright),
endalign
$$
and substitute these into the sum to get
$$
I_1(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$
and hence
$$
I(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$
as $x to infty$. Since $N$ was arbitrary, this is precisely the statement that $I(v)$ has the asymptotic expansion
$$
I(v) approx sum_k=0^infty fracg^(k)(0)x^k+1
$$
as $x to infty$.
answered Aug 22 at 5:03
Antonio Vargas
20.3k244110
Thank you, Antonio Vargas. I tried the v transformation earlier but did not proceed with the binomial expansion. The singularity at $v=1$ however prevents the direct application of Watson's lemma. I have modified your answer with an integration by parts to rid the singularity there. But it seems not to match the other answer though. I have not figured out why.
– Hans
Aug 22 at 8:46
@Hans Watson's lemma has no issue with the singularity at $v=1$. To apply it to an integral of the form $int_0^1 e^-xv g(v),dv$, Watson's lemma only requires (for example) $g(v)$ to be to be analytic at $v=0$ and for $int_0^1 |g(v)|,dv$ to exist. See the proof of Watson's lemma on the wikipedia page (which I wrote). I have reverted the edits you made to my answer. It does match the other answer (I double checked in Mathematica), so I'm not sure where you're going wrong.
– Antonio Vargas
Aug 23 at 0:17
Hmm, it's been a while since I wrote that proof, but in rereading it now it looks like it doesn't quite apply to the case in question. It can be modified to work on this case, however.
– Antonio Vargas
Aug 23 at 0:36
@Hans I've added a proof of Watson's lemma (only slightly modified from the one I linked) which applies directly to this specific question.
– Antonio Vargas
Aug 23 at 1:16
It looks good. Maybe you can incorporate your current more general proof into the Wikipedia article. Also, do you see why the integration by parts approach I took did not give the right answer?
– Hans
Aug 23 at 1:33
 |Â
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5
down vote
The integral can be solved in closed form in terms of the Dawson function.
$$f(x)=F(sqrtx)/sqrtx sim frac12x+frac14x^2+frac38x^3 +...$$
Mathematica was used to perform the asymptotic expansion for $s to infty.$
answered Aug 22 at 4:32
skbmoore
1,42229
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
The minimum of $1-u^2$ occurs at $u=1$, and near there we have $1-u^2 approx 2(1-u)$, which is linear in the quantity $1-u$. This suggests we make the change of variables $1-u^2 = v$ (which is linear in $v$), giving
$$
int_0^1 e^-x(1-u^2),du = frac12 int_0^1 e^-xv (1-v)^-1/2,dv.
$$
Then, following Watson's lemma, we can get the asymptotic expansion by expanding the subdominant term, $(1-v)^-1/2$, around $v=0$ and integrating term-by-term from $v=0$ to $v=infty$:
$$
beginalign
int_0^1 e^-x(1-u^2),du &= frac12 int_0^1 e^-xv (1-v)^-1/2,dv \
&approx frac12 sum_k=0^infty binom-1/2k (-1)^k int_0^infty e^-xv v^k,dv \
&= frac12 sum_k=0^infty binom-1/2k frac(-1)^k k!x^k+1
endalign
$$
as $x to infty$, which matches the series given in the other answer.
Let's prove the special case of Watson's lemma we use in this answer. We'll assume that $g(v)$ is analytic at $v=0$ and that $int_0^a lvert g(v) rvert,dv$ exists (these are certainly true for $g(v) = (1-v)^-1/2$), and show that
$$
I(v) = int_0^a e^-xvg(v),dv approx sum_k=0^infty fracg^(k)(0)x^k+1
$$
as $x to infty$. In other words, we will show that the asymptotic expansion for the integral can be obtained by expanding $g(v)$ in Taylor series around $v=0$ and integrating term-by-term.
Since $g(v)$ is analytic at $v=0$, there is a $delta in (0,a]$ such that $g^(k)(v)$ is analytic on $[0,delta]$ for all $k$. Further, Taylor's theorem tells us that for any positive integer $N$ and any $v in [0,delta]$, there is a $v^* in [0,v]$ for which
$$
g(v) = sum_k=0^N fracg^(k)(0)k! v^k + fracg^(N+1)(v^*)(N+1)! v^N+1. tag1
$$
We will split the integral $I(v)$ at this $delta$, and estimate each piece separately. To this end, we define
$$
I(v) = int_0^delta e^-xvg(v),dv + int_delta^a e^-xvg(v),dv = I_1(v) + I_2(v).
$$
We only need a rough estimate on $I_2(v)$:
$$
lvert I_2(v) rvert leq int_delta^a e^-xv lvert g(v) rvert ,dv leq e^-delta x int_0^a lvert g(v) rvert,dv,
$$
where we have assumed the last integral on the right is finite. Thus
$$
I(v) = I_1(v) + O(e^-delta x)
$$
as $x to infty$.
Now, from $(1)$ we have
$$
I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + frac1(N+1)! int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv.
$$
The last integral can be bounded by
$$
leftlvert int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv rightrvert leq left( sup_0 < v < delta leftlvert g^(N+1)(v) rightrvert right) int_0^infty e^-xv v^N+1,dv = fractextconst.x^N+2.
$$
Thus
$$
I_1(v) = I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + O!left(x^-N-2right)
$$
as $x to infty$. Finally we reattach the tails to the integrals in the sum,
$$
beginalign
int_0^delta e^-xv v^k ,dv &= int_0^infty e^-xv v^k,dv - int_delta^infty e^-xv v^k,dv \
&= frack!x^k+1 + O!left(e^-delta xright),
endalign
$$
and substitute these into the sum to get
$$
I_1(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$
and hence
$$
I(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$
as $x to infty$. Since $N$ was arbitrary, this is precisely the statement that $I(v)$ has the asymptotic expansion
$$
I(v) approx sum_k=0^infty fracg^(k)(0)x^k+1
$$
as $x to infty$.
answered Aug 22 at 5:03
Antonio Vargas
20.3k244110
Thank you, Antonio Vargas. I tried the v transformation earlier but did not proceed with the binomial expansion. The singularity at $v=1$ however prevents the direct application of Watson's lemma. I have modified your answer with an integration by parts to rid the singularity there. But it seems not to match the other answer though. I have not figured out why.
– Hans
Aug 22 at 8:46
@Hans Watson's lemma has no issue with the singularity at $v=1$. To apply it to an integral of the form $int_0^1 e^-xv g(v),dv$, Watson's lemma only requires (for example) $g(v)$ to be to be analytic at $v=0$ and for $int_0^1 |g(v)|,dv$ to exist. See the proof of Watson's lemma on the wikipedia page (which I wrote). I have reverted the edits you made to my answer. It does match the other answer (I double checked in Mathematica), so I'm not sure where you're going wrong.
– Antonio Vargas
Aug 23 at 0:17
Hmm, it's been a while since I wrote that proof, but in rereading it now it looks like it doesn't quite apply to the case in question. It can be modified to work on this case, however.
– Antonio Vargas
Aug 23 at 0:36
@Hans I've added a proof of Watson's lemma (only slightly modified from the one I linked) which applies directly to this specific question.
– Antonio Vargas
Aug 23 at 1:16
It looks good. Maybe you can incorporate your current more general proof into the Wikipedia article. Also, do you see why the integration by parts approach I took did not give the right answer?
– Hans
Aug 23 at 1:33
 |Â
show 4 more comments
up vote
6
down vote
accepted
The minimum of $1-u^2$ occurs at $u=1$, and near there we have $1-u^2 approx 2(1-u)$, which is linear in the quantity $1-u$. This suggests we make the change of variables $1-u^2 = v$ (which is linear in $v$), giving
$$
int_0^1 e^-x(1-u^2),du = frac12 int_0^1 e^-xv (1-v)^-1/2,dv.
$$
Then, following Watson's lemma, we can get the asymptotic expansion by expanding the subdominant term, $(1-v)^-1/2$, around $v=0$ and integrating term-by-term from $v=0$ to $v=infty$:
$$
beginalign
int_0^1 e^-x(1-u^2),du &= frac12 int_0^1 e^-xv (1-v)^-1/2,dv \
&approx frac12 sum_k=0^infty binom-1/2k (-1)^k int_0^infty e^-xv v^k,dv \
&= frac12 sum_k=0^infty binom-1/2k frac(-1)^k k!x^k+1
endalign
$$
as $x to infty$, which matches the series given in the other answer.
Let's prove the special case of Watson's lemma we use in this answer. We'll assume that $g(v)$ is analytic at $v=0$ and that $int_0^a lvert g(v) rvert,dv$ exists (these are certainly true for $g(v) = (1-v)^-1/2$), and show that
$$
I(v) = int_0^a e^-xvg(v),dv approx sum_k=0^infty fracg^(k)(0)x^k+1
$$
as $x to infty$. In other words, we will show that the asymptotic expansion for the integral can be obtained by expanding $g(v)$ in Taylor series around $v=0$ and integrating term-by-term.
Since $g(v)$ is analytic at $v=0$, there is a $delta in (0,a]$ such that $g^(k)(v)$ is analytic on $[0,delta]$ for all $k$. Further, Taylor's theorem tells us that for any positive integer $N$ and any $v in [0,delta]$, there is a $v^* in [0,v]$ for which
$$
g(v) = sum_k=0^N fracg^(k)(0)k! v^k + fracg^(N+1)(v^*)(N+1)! v^N+1. tag1
$$
We will split the integral $I(v)$ at this $delta$, and estimate each piece separately. To this end, we define
$$
I(v) = int_0^delta e^-xvg(v),dv + int_delta^a e^-xvg(v),dv = I_1(v) + I_2(v).
$$
We only need a rough estimate on $I_2(v)$:
$$
lvert I_2(v) rvert leq int_delta^a e^-xv lvert g(v) rvert ,dv leq e^-delta x int_0^a lvert g(v) rvert,dv,
$$
where we have assumed the last integral on the right is finite. Thus
$$
I(v) = I_1(v) + O(e^-delta x)
$$
as $x to infty$.
Now, from $(1)$ we have
$$
I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + frac1(N+1)! int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv.
$$
The last integral can be bounded by
$$
leftlvert int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv rightrvert leq left( sup_0 < v < delta leftlvert g^(N+1)(v) rightrvert right) int_0^infty e^-xv v^N+1,dv = fractextconst.x^N+2.
$$
Thus
$$
I_1(v) = I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + O!left(x^-N-2right)
$$
as $x to infty$. Finally we reattach the tails to the integrals in the sum,
$$
beginalign
int_0^delta e^-xv v^k ,dv &= int_0^infty e^-xv v^k,dv - int_delta^infty e^-xv v^k,dv \
&= frack!x^k+1 + O!left(e^-delta xright),
endalign
$$
and substitute these into the sum to get
$$
I_1(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$
and hence
$$
I(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$
as $x to infty$. Since $N$ was arbitrary, this is precisely the statement that $I(v)$ has the asymptotic expansion
$$
I(v) approx sum_k=0^infty fracg^(k)(0)x^k+1
$$
as $x to infty$.
answered Aug 22 at 5:03
Antonio Vargas
20.3k244110
Thank you, Antonio Vargas. I tried the v transformation earlier but did not proceed with the binomial expansion. The singularity at $v=1$ however prevents the direct application of Watson's lemma. I have modified your answer with an integration by parts to rid the singularity there. But it seems not to match the other answer though. I have not figured out why.
– Hans
Aug 22 at 8:46
@Hans Watson's lemma has no issue with the singularity at $v=1$. To apply it to an integral of the form $int_0^1 e^-xv g(v),dv$, Watson's lemma only requires (for example) $g(v)$ to be to be analytic at $v=0$ and for $int_0^1 |g(v)|,dv$ to exist. See the proof of Watson's lemma on the wikipedia page (which I wrote). I have reverted the edits you made to my answer. It does match the other answer (I double checked in Mathematica), so I'm not sure where you're going wrong.
– Antonio Vargas
Aug 23 at 0:17
Hmm, it's been a while since I wrote that proof, but in rereading it now it looks like it doesn't quite apply to the case in question. It can be modified to work on this case, however.
– Antonio Vargas
Aug 23 at 0:36
@Hans I've added a proof of Watson's lemma (only slightly modified from the one I linked) which applies directly to this specific question.
– Antonio Vargas
Aug 23 at 1:16
It looks good. Maybe you can incorporate your current more general proof into the Wikipedia article. Also, do you see why the integration by parts approach I took did not give the right answer?
– Hans
Aug 23 at 1:33
 |Â
show 4 more comments
up vote
6
down vote
accepted
up vote
6
down vote
accepted
The minimum of $1-u^2$ occurs at $u=1$, and near there we have $1-u^2 approx 2(1-u)$, which is linear in the quantity $1-u$. This suggests we make the change of variables $1-u^2 = v$ (which is linear in $v$), giving
$$
int_0^1 e^-x(1-u^2),du = frac12 int_0^1 e^-xv (1-v)^-1/2,dv.
$$
Then, following Watson's lemma, we can get the asymptotic expansion by expanding the subdominant term, $(1-v)^-1/2$, around $v=0$ and integrating term-by-term from $v=0$ to $v=infty$:
$$
beginalign
int_0^1 e^-x(1-u^2),du &= frac12 int_0^1 e^-xv (1-v)^-1/2,dv \
&approx frac12 sum_k=0^infty binom-1/2k (-1)^k int_0^infty e^-xv v^k,dv \
&= frac12 sum_k=0^infty binom-1/2k frac(-1)^k k!x^k+1
endalign
$$
as $x to infty$, which matches the series given in the other answer.
Let's prove the special case of Watson's lemma we use in this answer. We'll assume that $g(v)$ is analytic at $v=0$ and that $int_0^a lvert g(v) rvert,dv$ exists (these are certainly true for $g(v) = (1-v)^-1/2$), and show that
$$
I(v) = int_0^a e^-xvg(v),dv approx sum_k=0^infty fracg^(k)(0)x^k+1
$$
as $x to infty$. In other words, we will show that the asymptotic expansion for the integral can be obtained by expanding $g(v)$ in Taylor series around $v=0$ and integrating term-by-term.
Since $g(v)$ is analytic at $v=0$, there is a $delta in (0,a]$ such that $g^(k)(v)$ is analytic on $[0,delta]$ for all $k$. Further, Taylor's theorem tells us that for any positive integer $N$ and any $v in [0,delta]$, there is a $v^* in [0,v]$ for which
$$
g(v) = sum_k=0^N fracg^(k)(0)k! v^k + fracg^(N+1)(v^*)(N+1)! v^N+1. tag1
$$
We will split the integral $I(v)$ at this $delta$, and estimate each piece separately. To this end, we define
$$
I(v) = int_0^delta e^-xvg(v),dv + int_delta^a e^-xvg(v),dv = I_1(v) + I_2(v).
$$
We only need a rough estimate on $I_2(v)$:
$$
lvert I_2(v) rvert leq int_delta^a e^-xv lvert g(v) rvert ,dv leq e^-delta x int_0^a lvert g(v) rvert,dv,
$$
where we have assumed the last integral on the right is finite. Thus
$$
I(v) = I_1(v) + O(e^-delta x)
$$
as $x to infty$.
Now, from $(1)$ we have
$$
I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + frac1(N+1)! int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv.
$$
The last integral can be bounded by
$$
leftlvert int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv rightrvert leq left( sup_0 < v < delta leftlvert g^(N+1)(v) rightrvert right) int_0^infty e^-xv v^N+1,dv = fractextconst.x^N+2.
$$
Thus
$$
I_1(v) = I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + O!left(x^-N-2right)
$$
as $x to infty$. Finally we reattach the tails to the integrals in the sum,
$$
beginalign
int_0^delta e^-xv v^k ,dv &= int_0^infty e^-xv v^k,dv - int_delta^infty e^-xv v^k,dv \
&= frack!x^k+1 + O!left(e^-delta xright),
endalign
$$
and substitute these into the sum to get
$$
I_1(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$
and hence
$$
I(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$
as $x to infty$. Since $N$ was arbitrary, this is precisely the statement that $I(v)$ has the asymptotic expansion
$$
I(v) approx sum_k=0^infty fracg^(k)(0)x^k+1
$$
as $x to infty$.
answered Aug 22 at 5:03
Antonio Vargas
20.3k244110
The minimum of $1-u^2$ occurs at $u=1$, and near there we have $1-u^2 approx 2(1-u)$, which is linear in the quantity $1-u$. This suggests we make the change of variables $1-u^2 = v$ (which is linear in $v$), giving
$$
int_0^1 e^-x(1-u^2),du = frac12 int_0^1 e^-xv (1-v)^-1/2,dv.
$$
Then, following Watson's lemma, we can get the asymptotic expansion by expanding the subdominant term, $(1-v)^-1/2$, around $v=0$ and integrating term-by-term from $v=0$ to $v=infty$:
$$
beginalign
int_0^1 e^-x(1-u^2),du &= frac12 int_0^1 e^-xv (1-v)^-1/2,dv \
&approx frac12 sum_k=0^infty binom-1/2k (-1)^k int_0^infty e^-xv v^k,dv \
&= frac12 sum_k=0^infty binom-1/2k frac(-1)^k k!x^k+1
endalign
$$
as $x to infty$, which matches the series given in the other answer.
Let's prove the special case of Watson's lemma we use in this answer. We'll assume that $g(v)$ is analytic at $v=0$ and that $int_0^a lvert g(v) rvert,dv$ exists (these are certainly true for $g(v) = (1-v)^-1/2$), and show that
$$
I(v) = int_0^a e^-xvg(v),dv approx sum_k=0^infty fracg^(k)(0)x^k+1
$$
as $x to infty$. In other words, we will show that the asymptotic expansion for the integral can be obtained by expanding $g(v)$ in Taylor series around $v=0$ and integrating term-by-term.
Since $g(v)$ is analytic at $v=0$, there is a $delta in (0,a]$ such that $g^(k)(v)$ is analytic on $[0,delta]$ for all $k$. Further, Taylor's theorem tells us that for any positive integer $N$ and any $v in [0,delta]$, there is a $v^* in [0,v]$ for which
$$
g(v) = sum_k=0^N fracg^(k)(0)k! v^k + fracg^(N+1)(v^*)(N+1)! v^N+1. tag1
$$
We will split the integral $I(v)$ at this $delta$, and estimate each piece separately. To this end, we define
$$
I(v) = int_0^delta e^-xvg(v),dv + int_delta^a e^-xvg(v),dv = I_1(v) + I_2(v).
$$
We only need a rough estimate on $I_2(v)$:
$$
lvert I_2(v) rvert leq int_delta^a e^-xv lvert g(v) rvert ,dv leq e^-delta x int_0^a lvert g(v) rvert,dv,
$$
where we have assumed the last integral on the right is finite. Thus
$$
I(v) = I_1(v) + O(e^-delta x)
$$
as $x to infty$.
Now, from $(1)$ we have
$$
I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + frac1(N+1)! int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv.
$$
The last integral can be bounded by
$$
leftlvert int_0^delta e^-xv g^(N+1)(v^*) v^N+1,dv rightrvert leq left( sup_0 < v < delta leftlvert g^(N+1)(v) rightrvert right) int_0^infty e^-xv v^N+1,dv = fractextconst.x^N+2.
$$
Thus
$$
I_1(v) = I_1(v) = sum_k=0^N fracg^(k)(0)k! int_0^delta e^-xv v^k ,dv + O!left(x^-N-2right)
$$
as $x to infty$. Finally we reattach the tails to the integrals in the sum,
$$
beginalign
int_0^delta e^-xv v^k ,dv &= int_0^infty e^-xv v^k,dv - int_delta^infty e^-xv v^k,dv \
&= frack!x^k+1 + O!left(e^-delta xright),
endalign
$$
and substitute these into the sum to get
$$
I_1(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$
and hence
$$
I(v) = sum_k=0^N fracg^(k)(0)x^k+1 + O!left(x^-N-2right)
$$
as $x to infty$. Since $N$ was arbitrary, this is precisely the statement that $I(v)$ has the asymptotic expansion
$$
I(v) approx sum_k=0^infty fracg^(k)(0)x^k+1
$$
as $x to infty$.
answered Aug 22 at 5:03
Antonio Vargas
20.3k244110
edited Aug 23 at 1:56
answered Aug 22 at 5:03
Antonio Vargas
20.3k244110
answered Aug 22 at 5:03
Antonio Vargas
20.3k244110
answered Aug 22 at 5:03
Antonio Vargas
20.3k244110
20.3k244110
Thank you, Antonio Vargas. I tried the v transformation earlier but did not proceed with the binomial expansion. The singularity at $v=1$ however prevents the direct application of Watson's lemma. I have modified your answer with an integration by parts to rid the singularity there. But it seems not to match the other answer though. I have not figured out why.
– Hans
Aug 22 at 8:46
@Hans Watson's lemma has no issue with the singularity at $v=1$. To apply it to an integral of the form $int_0^1 e^-xv g(v),dv$, Watson's lemma only requires (for example) $g(v)$ to be to be analytic at $v=0$ and for $int_0^1 |g(v)|,dv$ to exist. See the proof of Watson's lemma on the wikipedia page (which I wrote). I have reverted the edits you made to my answer. It does match the other answer (I double checked in Mathematica), so I'm not sure where you're going wrong.
– Antonio Vargas
Aug 23 at 0:17
Hmm, it's been a while since I wrote that proof, but in rereading it now it looks like it doesn't quite apply to the case in question. It can be modified to work on this case, however.
– Antonio Vargas
Aug 23 at 0:36
@Hans I've added a proof of Watson's lemma (only slightly modified from the one I linked) which applies directly to this specific question.
– Antonio Vargas
Aug 23 at 1:16
It looks good. Maybe you can incorporate your current more general proof into the Wikipedia article. Also, do you see why the integration by parts approach I took did not give the right answer?
– Hans
Aug 23 at 1:33
 |Â
show 4 more comments
Thank you, Antonio Vargas. I tried the v transformation earlier but did not proceed with the binomial expansion. The singularity at $v=1$ however prevents the direct application of Watson's lemma. I have modified your answer with an integration by parts to rid the singularity there. But it seems not to match the other answer though. I have not figured out why.
– Hans
Aug 22 at 8:46
@Hans Watson's lemma has no issue with the singularity at $v=1$. To apply it to an integral of the form $int_0^1 e^-xv g(v),dv$, Watson's lemma only requires (for example) $g(v)$ to be to be analytic at $v=0$ and for $int_0^1 |g(v)|,dv$ to exist. See the proof of Watson's lemma on the wikipedia page (which I wrote). I have reverted the edits you made to my answer. It does match the other answer (I double checked in Mathematica), so I'm not sure where you're going wrong.
– Antonio Vargas
Aug 23 at 0:17
Hmm, it's been a while since I wrote that proof, but in rereading it now it looks like it doesn't quite apply to the case in question. It can be modified to work on this case, however.
– Antonio Vargas
Aug 23 at 0:36
@Hans I've added a proof of Watson's lemma (only slightly modified from the one I linked) which applies directly to this specific question.
– Antonio Vargas
Aug 23 at 1:16
It looks good. Maybe you can incorporate your current more general proof into the Wikipedia article. Also, do you see why the integration by parts approach I took did not give the right answer?
– Hans
Aug 23 at 1:33
Thank you, Antonio Vargas. I tried the v transformation earlier but did not proceed with the binomial expansion. The singularity at $v=1$ however prevents the direct application of Watson's lemma. I have modified your answer with an integration by parts to rid the singularity there. But it seems not to match the other answer though. I have not figured out why.
– Hans
Aug 22 at 8:46
Thank you, Antonio Vargas. I tried the v transformation earlier but did not proceed with the binomial expansion. The singularity at $v=1$ however prevents the direct application of Watson's lemma. I have modified your answer with an integration by parts to rid the singularity there. But it seems not to match the other answer though. I have not figured out why.
– Hans
Aug 22 at 8:46
@Hans Watson's lemma has no issue with the singularity at $v=1$. To apply it to an integral of the form $int_0^1 e^-xv g(v),dv$, Watson's lemma only requires (for example) $g(v)$ to be to be analytic at $v=0$ and for $int_0^1 |g(v)|,dv$ to exist. See the proof of Watson's lemma on the wikipedia page (which I wrote). I have reverted the edits you made to my answer. It does match the other answer (I double checked in Mathematica), so I'm not sure where you're going wrong.
– Antonio Vargas
Aug 23 at 0:17
@Hans Watson's lemma has no issue with the singularity at $v=1$. To apply it to an integral of the form $int_0^1 e^-xv g(v),dv$, Watson's lemma only requires (for example) $g(v)$ to be to be analytic at $v=0$ and for $int_0^1 |g(v)|,dv$ to exist. See the proof of Watson's lemma on the wikipedia page (which I wrote). I have reverted the edits you made to my answer. It does match the other answer (I double checked in Mathematica), so I'm not sure where you're going wrong.
– Antonio Vargas
Aug 23 at 0:17
Hmm, it's been a while since I wrote that proof, but in rereading it now it looks like it doesn't quite apply to the case in question. It can be modified to work on this case, however.
– Antonio Vargas
Aug 23 at 0:36
Hmm, it's been a while since I wrote that proof, but in rereading it now it looks like it doesn't quite apply to the case in question. It can be modified to work on this case, however.
– Antonio Vargas
Aug 23 at 0:36
@Hans I've added a proof of Watson's lemma (only slightly modified from the one I linked) which applies directly to this specific question.
– Antonio Vargas
Aug 23 at 1:16
@Hans I've added a proof of Watson's lemma (only slightly modified from the one I linked) which applies directly to this specific question.
– Antonio Vargas
Aug 23 at 1:16
It looks good. Maybe you can incorporate your current more general proof into the Wikipedia article. Also, do you see why the integration by parts approach I took did not give the right answer?
– Hans
Aug 23 at 1:33
It looks good. Maybe you can incorporate your current more general proof into the Wikipedia article. Also, do you see why the integration by parts approach I took did not give the right answer?
– Hans
Aug 23 at 1:33
 |Â
show 4 more comments
up vote
5
down vote
The integral can be solved in closed form in terms of the Dawson function.
$$f(x)=F(sqrtx)/sqrtx sim frac12x+frac14x^2+frac38x^3 +...$$
Mathematica was used to perform the asymptotic expansion for $s to infty.$
answered Aug 22 at 4:32
skbmoore
1,42229
add a comment |Â
up vote
5
down vote
The integral can be solved in closed form in terms of the Dawson function.
$$f(x)=F(sqrtx)/sqrtx sim frac12x+frac14x^2+frac38x^3 +...$$
Mathematica was used to perform the asymptotic expansion for $s to infty.$
answered Aug 22 at 4:32
skbmoore
1,42229
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The integral can be solved in closed form in terms of the Dawson function.
$$f(x)=F(sqrtx)/sqrtx sim frac12x+frac14x^2+frac38x^3 +...$$
Mathematica was used to perform the asymptotic expansion for $s to infty.$
answered Aug 22 at 4:32
skbmoore
1,42229
The integral can be solved in closed form in terms of the Dawson function.
$$f(x)=F(sqrtx)/sqrtx sim frac12x+frac14x^2+frac38x^3 +...$$
Mathematica was used to perform the asymptotic expansion for $s to infty.$
answered Aug 22 at 4:32
skbmoore
1,42229
answered Aug 22 at 4:32
skbmoore
1,42229
answered Aug 22 at 4:32
skbmoore
1,42229
answered Aug 22 at 4:32
skbmoore
1,42229
1,42229
add a comment |Â
add a comment |Â
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