Mixing
Integral of Binomial Coefficient
Clash Royale CLAN TAG#URR8PPP
up vote
13
down vote
favorite
We all know the famous theorem that:
$$sum_i=1^nbinomni=2^n$$
This theorem got me wondering about a similar formula - the properties of the following function:
$$I(n)=int_0^n binomnk,,mathrmdk$$
where $n$ is any positive integer and the definition of binomial coefficient is "extended" by way of gamma functions (i.e the integrand is really $fracGamma(n+1)Gamma(k+1)Gamma(n-k+1)$).
What I found, experimentally, is pretty cool. It seems that the following is true:
$$I(n)=frac2pi sum_i=1^n binomnioperatornameSinInt(pi i)$$
Where $operatornameSinInt(x)$ is the Sine Integral, or $int_0^x fracsin ttdt$.
To me, this is quite interesting as the Sine Integral tends to $pi/2$ so the above formula will tend to $2^n$, so the integral is just the sum with some error term. But how would I go about proving it?
calculus integration definite-integrals
asked 2 hours ago
TreFox
1,89011033
add a comment |Â
up vote
13
down vote
favorite
We all know the famous theorem that:
$$sum_i=1^nbinomni=2^n$$
This theorem got me wondering about a similar formula - the properties of the following function:
$$I(n)=int_0^n binomnk,,mathrmdk$$
where $n$ is any positive integer and the definition of binomial coefficient is "extended" by way of gamma functions (i.e the integrand is really $fracGamma(n+1)Gamma(k+1)Gamma(n-k+1)$).
What I found, experimentally, is pretty cool. It seems that the following is true:
$$I(n)=frac2pi sum_i=1^n binomnioperatornameSinInt(pi i)$$
Where $operatornameSinInt(x)$ is the Sine Integral, or $int_0^x fracsin ttdt$.
To me, this is quite interesting as the Sine Integral tends to $pi/2$ so the above formula will tend to $2^n$, so the integral is just the sum with some error term. But how would I go about proving it?
calculus integration definite-integrals
asked 2 hours ago
TreFox
1,89011033
2
It might be interesting to mention that $$int_0^1 binomx n dx= G_n$$ where $G_n$ are Gregory coefficients.
– Dahaka
2 hours ago
Since $n$ does not change as $k$ goes from $0$ to $n,$ one has $displaystyle I(n) = n!int_0^n fracmathrm dkk!(n-k)!. qquad$
– Michael Hardy
8 mins ago
How did you find it experimentally? Using some tools?
– Dilworth
29 secs ago
add a comment |Â
up vote
13
down vote
favorite
up vote
13
down vote
favorite
We all know the famous theorem that:
$$sum_i=1^nbinomni=2^n$$
This theorem got me wondering about a similar formula - the properties of the following function:
$$I(n)=int_0^n binomnk,,mathrmdk$$
where $n$ is any positive integer and the definition of binomial coefficient is "extended" by way of gamma functions (i.e the integrand is really $fracGamma(n+1)Gamma(k+1)Gamma(n-k+1)$).
What I found, experimentally, is pretty cool. It seems that the following is true:
$$I(n)=frac2pi sum_i=1^n binomnioperatornameSinInt(pi i)$$
Where $operatornameSinInt(x)$ is the Sine Integral, or $int_0^x fracsin ttdt$.
To me, this is quite interesting as the Sine Integral tends to $pi/2$ so the above formula will tend to $2^n$, so the integral is just the sum with some error term. But how would I go about proving it?
calculus integration definite-integrals
asked 2 hours ago
TreFox
1,89011033
We all know the famous theorem that:
$$sum_i=1^nbinomni=2^n$$
This theorem got me wondering about a similar formula - the properties of the following function:
$$I(n)=int_0^n binomnk,,mathrmdk$$
where $n$ is any positive integer and the definition of binomial coefficient is "extended" by way of gamma functions (i.e the integrand is really $fracGamma(n+1)Gamma(k+1)Gamma(n-k+1)$).
What I found, experimentally, is pretty cool. It seems that the following is true:
$$I(n)=frac2pi sum_i=1^n binomnioperatornameSinInt(pi i)$$
Where $operatornameSinInt(x)$ is the Sine Integral, or $int_0^x fracsin ttdt$.
To me, this is quite interesting as the Sine Integral tends to $pi/2$ so the above formula will tend to $2^n$, so the integral is just the sum with some error term. But how would I go about proving it?
calculus integration definite-integrals
calculus integration definite-integrals
asked 2 hours ago
TreFox
1,89011033
asked 2 hours ago
TreFox
1,89011033
asked 2 hours ago
TreFox
1,89011033
asked 2 hours ago
TreFox
1,89011033
asked 2 hours ago
TreFox
1,89011033
1,89011033
2
It might be interesting to mention that $$int_0^1 binomx n dx= G_n$$ where $G_n$ are Gregory coefficients.
– Dahaka
2 hours ago
Since $n$ does not change as $k$ goes from $0$ to $n,$ one has $displaystyle I(n) = n!int_0^n fracmathrm dkk!(n-k)!. qquad$
– Michael Hardy
8 mins ago
How did you find it experimentally? Using some tools?
– Dilworth
29 secs ago
add a comment |Â
2
It might be interesting to mention that $$int_0^1 binomx n dx= G_n$$ where $G_n$ are Gregory coefficients.
– Dahaka
2 hours ago
Since $n$ does not change as $k$ goes from $0$ to $n,$ one has $displaystyle I(n) = n!int_0^n fracmathrm dkk!(n-k)!. qquad$
– Michael Hardy
8 mins ago
How did you find it experimentally? Using some tools?
– Dilworth
29 secs ago
2
2
It might be interesting to mention that $$int_0^1 binomx n dx= G_n$$ where $G_n$ are Gregory coefficients.
– Dahaka
2 hours ago
It might be interesting to mention that $$int_0^1 binomx n dx= G_n$$ where $G_n$ are Gregory coefficients.
– Dahaka
2 hours ago
Since $n$ does not change as $k$ goes from $0$ to $n,$ one has $displaystyle I(n) = n!int_0^n fracmathrm dkk!(n-k)!. qquad$
– Michael Hardy
8 mins ago
Since $n$ does not change as $k$ goes from $0$ to $n,$ one has $displaystyle I(n) = n!int_0^n fracmathrm dkk!(n-k)!. qquad$
– Michael Hardy
8 mins ago
How did you find it experimentally? Using some tools?
– Dilworth
29 secs ago
How did you find it experimentally? Using some tools?
– Dilworth
29 secs ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
Properties of the $Gamma$-function, together with partial fraction expansion, do the trick. We have
$$Gamma(x+1)Gamma(n-x+1)=xGamma(x)Gamma(1-x)prod_k=1^n(k-x)=frac(-1)^npisinpi xprod_k=0^n(x-k),$$
so our integral is $I_n=displaystylefrac(-1)^n n!piint_0^nfracsinpi x,mathrmdxprod_k=0^n(x-k)$. Doing fraction expansion, we have
$$prod_k=0^n(x-k)^-1=sum_k=0^nfraca_kx-k,quad a_m=prod_substack0leq kleq n\kneq m(m-k)^-1=frac(-1)^n-mm!(n-m)!$$
(say, multiplying by $x-m$ and letting $xto m$). Thus we get
$$I_n=frac1piint_0^nsum_k=0^n(-1)^kbinomnkfracsinpi xx-k,mathrmdx=frac1pisum_k=0^nbinomnkint_-kpi^(n-k)pifracsin tt,mathrmdt.$$
Simplification of this, using the sine integral function, gives the expected result.
answered 1 hour ago
metamorphy
1,40410
add a comment |Â
up vote
1
down vote
When I was your age (15 or so), I also played around with a very similar integral. My take was slightly different.
First of all, for suitable real numbers $s$ (say with positive real part) one actually has the sum
$$sum_-infty^infty binomsk = 2^s.$$
So a more natural integral to consider might be
$$I(s):=int_-infty^infty fracGamma(s+1)Gamma(x+1) Gamma(s+1-x) , dx.$$
You can prove $I(n) = 2^n$ for non-negative integers $n$. The point here is that one can use the reflection formula for the $Gamma$ function. For example, in the easiest case when $s = 0$, the integral becomes
beginalign
& int_-infty^infty frac1Gamma(x+1) Gamma(1-x) , dx \[10pt]
= & int_-infty^infty frac1x Gamma(x) Gamma(1-x) , dx \[10pt]
= & int_-infty^infty fracsin(pi x)x pi , dx = 1.
endalign
For larger integers $n$, you can do pretty much the same thing and then use partial fractions. The same argument will work with your integral, except now the sin integral will go from $0$ to $n$ rather than $infty$, and hence you pick up the corresponding functions.
(and looks like someone has done that).
This doesn't work for general $s$, however. It took me quite a while to work out, but eventually I found a few different arguments to prove that $I(s) = 2^s$, including a fairly clean proof by contour integration.
Of course, all this was in the days before the internet, so I had to figure it out on my own by thinking about it (even though it was surely known before and also not that hard in the end --- there are very few truly deep definite integrals). I feel a little sad that you can just ask the question and someone will come and answer it.
edited 6 mins ago
Michael Hardy
206k23187466
answered 11 mins ago
Sad Clown
111
New contributor
Sad Clown is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thank you so much for the story to go along with your answer. Rest assured I spend much more time playing around with math on my own than I spend asking about it here! I just haven't actually had a formal calculus class so integrals with special functions always tend to scare me off :) But this was too cool to ignore. I'm currently trying to teach myself contour integration, so I'll search for the proof you mention! Once again, thank you.
– TreFox
5 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Properties of the $Gamma$-function, together with partial fraction expansion, do the trick. We have
$$Gamma(x+1)Gamma(n-x+1)=xGamma(x)Gamma(1-x)prod_k=1^n(k-x)=frac(-1)^npisinpi xprod_k=0^n(x-k),$$
so our integral is $I_n=displaystylefrac(-1)^n n!piint_0^nfracsinpi x,mathrmdxprod_k=0^n(x-k)$. Doing fraction expansion, we have
$$prod_k=0^n(x-k)^-1=sum_k=0^nfraca_kx-k,quad a_m=prod_substack0leq kleq n\kneq m(m-k)^-1=frac(-1)^n-mm!(n-m)!$$
(say, multiplying by $x-m$ and letting $xto m$). Thus we get
$$I_n=frac1piint_0^nsum_k=0^n(-1)^kbinomnkfracsinpi xx-k,mathrmdx=frac1pisum_k=0^nbinomnkint_-kpi^(n-k)pifracsin tt,mathrmdt.$$
Simplification of this, using the sine integral function, gives the expected result.
answered 1 hour ago
metamorphy
1,40410
add a comment |Â
up vote
4
down vote
Properties of the $Gamma$-function, together with partial fraction expansion, do the trick. We have
$$Gamma(x+1)Gamma(n-x+1)=xGamma(x)Gamma(1-x)prod_k=1^n(k-x)=frac(-1)^npisinpi xprod_k=0^n(x-k),$$
so our integral is $I_n=displaystylefrac(-1)^n n!piint_0^nfracsinpi x,mathrmdxprod_k=0^n(x-k)$. Doing fraction expansion, we have
$$prod_k=0^n(x-k)^-1=sum_k=0^nfraca_kx-k,quad a_m=prod_substack0leq kleq n\kneq m(m-k)^-1=frac(-1)^n-mm!(n-m)!$$
(say, multiplying by $x-m$ and letting $xto m$). Thus we get
$$I_n=frac1piint_0^nsum_k=0^n(-1)^kbinomnkfracsinpi xx-k,mathrmdx=frac1pisum_k=0^nbinomnkint_-kpi^(n-k)pifracsin tt,mathrmdt.$$
Simplification of this, using the sine integral function, gives the expected result.
answered 1 hour ago
metamorphy
1,40410
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Properties of the $Gamma$-function, together with partial fraction expansion, do the trick. We have
$$Gamma(x+1)Gamma(n-x+1)=xGamma(x)Gamma(1-x)prod_k=1^n(k-x)=frac(-1)^npisinpi xprod_k=0^n(x-k),$$
so our integral is $I_n=displaystylefrac(-1)^n n!piint_0^nfracsinpi x,mathrmdxprod_k=0^n(x-k)$. Doing fraction expansion, we have
$$prod_k=0^n(x-k)^-1=sum_k=0^nfraca_kx-k,quad a_m=prod_substack0leq kleq n\kneq m(m-k)^-1=frac(-1)^n-mm!(n-m)!$$
(say, multiplying by $x-m$ and letting $xto m$). Thus we get
$$I_n=frac1piint_0^nsum_k=0^n(-1)^kbinomnkfracsinpi xx-k,mathrmdx=frac1pisum_k=0^nbinomnkint_-kpi^(n-k)pifracsin tt,mathrmdt.$$
Simplification of this, using the sine integral function, gives the expected result.
answered 1 hour ago
metamorphy
1,40410
Properties of the $Gamma$-function, together with partial fraction expansion, do the trick. We have
$$Gamma(x+1)Gamma(n-x+1)=xGamma(x)Gamma(1-x)prod_k=1^n(k-x)=frac(-1)^npisinpi xprod_k=0^n(x-k),$$
so our integral is $I_n=displaystylefrac(-1)^n n!piint_0^nfracsinpi x,mathrmdxprod_k=0^n(x-k)$. Doing fraction expansion, we have
$$prod_k=0^n(x-k)^-1=sum_k=0^nfraca_kx-k,quad a_m=prod_substack0leq kleq n\kneq m(m-k)^-1=frac(-1)^n-mm!(n-m)!$$
(say, multiplying by $x-m$ and letting $xto m$). Thus we get
$$I_n=frac1piint_0^nsum_k=0^n(-1)^kbinomnkfracsinpi xx-k,mathrmdx=frac1pisum_k=0^nbinomnkint_-kpi^(n-k)pifracsin tt,mathrmdt.$$
Simplification of this, using the sine integral function, gives the expected result.
answered 1 hour ago
metamorphy
1,40410
edited 1 hour ago
answered 1 hour ago
metamorphy
1,40410
answered 1 hour ago
metamorphy
1,40410
answered 1 hour ago
metamorphy
1,40410
1,40410
add a comment |Â
add a comment |Â
up vote
1
down vote
When I was your age (15 or so), I also played around with a very similar integral. My take was slightly different.
First of all, for suitable real numbers $s$ (say with positive real part) one actually has the sum
$$sum_-infty^infty binomsk = 2^s.$$
So a more natural integral to consider might be
$$I(s):=int_-infty^infty fracGamma(s+1)Gamma(x+1) Gamma(s+1-x) , dx.$$
You can prove $I(n) = 2^n$ for non-negative integers $n$. The point here is that one can use the reflection formula for the $Gamma$ function. For example, in the easiest case when $s = 0$, the integral becomes
beginalign
& int_-infty^infty frac1Gamma(x+1) Gamma(1-x) , dx \[10pt]
= & int_-infty^infty frac1x Gamma(x) Gamma(1-x) , dx \[10pt]
= & int_-infty^infty fracsin(pi x)x pi , dx = 1.
endalign
For larger integers $n$, you can do pretty much the same thing and then use partial fractions. The same argument will work with your integral, except now the sin integral will go from $0$ to $n$ rather than $infty$, and hence you pick up the corresponding functions.
(and looks like someone has done that).
This doesn't work for general $s$, however. It took me quite a while to work out, but eventually I found a few different arguments to prove that $I(s) = 2^s$, including a fairly clean proof by contour integration.
Of course, all this was in the days before the internet, so I had to figure it out on my own by thinking about it (even though it was surely known before and also not that hard in the end --- there are very few truly deep definite integrals). I feel a little sad that you can just ask the question and someone will come and answer it.
edited 6 mins ago
Michael Hardy
206k23187466
answered 11 mins ago
Sad Clown
111
New contributor
Sad Clown is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thank you so much for the story to go along with your answer. Rest assured I spend much more time playing around with math on my own than I spend asking about it here! I just haven't actually had a formal calculus class so integrals with special functions always tend to scare me off :) But this was too cool to ignore. I'm currently trying to teach myself contour integration, so I'll search for the proof you mention! Once again, thank you.
– TreFox
5 mins ago
add a comment |Â
up vote
1
down vote
When I was your age (15 or so), I also played around with a very similar integral. My take was slightly different.
First of all, for suitable real numbers $s$ (say with positive real part) one actually has the sum
$$sum_-infty^infty binomsk = 2^s.$$
So a more natural integral to consider might be
$$I(s):=int_-infty^infty fracGamma(s+1)Gamma(x+1) Gamma(s+1-x) , dx.$$
You can prove $I(n) = 2^n$ for non-negative integers $n$. The point here is that one can use the reflection formula for the $Gamma$ function. For example, in the easiest case when $s = 0$, the integral becomes
beginalign
& int_-infty^infty frac1Gamma(x+1) Gamma(1-x) , dx \[10pt]
= & int_-infty^infty frac1x Gamma(x) Gamma(1-x) , dx \[10pt]
= & int_-infty^infty fracsin(pi x)x pi , dx = 1.
endalign
For larger integers $n$, you can do pretty much the same thing and then use partial fractions. The same argument will work with your integral, except now the sin integral will go from $0$ to $n$ rather than $infty$, and hence you pick up the corresponding functions.
(and looks like someone has done that).
This doesn't work for general $s$, however. It took me quite a while to work out, but eventually I found a few different arguments to prove that $I(s) = 2^s$, including a fairly clean proof by contour integration.
Of course, all this was in the days before the internet, so I had to figure it out on my own by thinking about it (even though it was surely known before and also not that hard in the end --- there are very few truly deep definite integrals). I feel a little sad that you can just ask the question and someone will come and answer it.
edited 6 mins ago
Michael Hardy
206k23187466
answered 11 mins ago
Sad Clown
111
New contributor
Sad Clown is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thank you so much for the story to go along with your answer. Rest assured I spend much more time playing around with math on my own than I spend asking about it here! I just haven't actually had a formal calculus class so integrals with special functions always tend to scare me off :) But this was too cool to ignore. I'm currently trying to teach myself contour integration, so I'll search for the proof you mention! Once again, thank you.
– TreFox
5 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
When I was your age (15 or so), I also played around with a very similar integral. My take was slightly different.
First of all, for suitable real numbers $s$ (say with positive real part) one actually has the sum
$$sum_-infty^infty binomsk = 2^s.$$
So a more natural integral to consider might be
$$I(s):=int_-infty^infty fracGamma(s+1)Gamma(x+1) Gamma(s+1-x) , dx.$$
You can prove $I(n) = 2^n$ for non-negative integers $n$. The point here is that one can use the reflection formula for the $Gamma$ function. For example, in the easiest case when $s = 0$, the integral becomes
beginalign
& int_-infty^infty frac1Gamma(x+1) Gamma(1-x) , dx \[10pt]
= & int_-infty^infty frac1x Gamma(x) Gamma(1-x) , dx \[10pt]
= & int_-infty^infty fracsin(pi x)x pi , dx = 1.
endalign
For larger integers $n$, you can do pretty much the same thing and then use partial fractions. The same argument will work with your integral, except now the sin integral will go from $0$ to $n$ rather than $infty$, and hence you pick up the corresponding functions.
(and looks like someone has done that).
This doesn't work for general $s$, however. It took me quite a while to work out, but eventually I found a few different arguments to prove that $I(s) = 2^s$, including a fairly clean proof by contour integration.
Of course, all this was in the days before the internet, so I had to figure it out on my own by thinking about it (even though it was surely known before and also not that hard in the end --- there are very few truly deep definite integrals). I feel a little sad that you can just ask the question and someone will come and answer it.
edited 6 mins ago
Michael Hardy
206k23187466
answered 11 mins ago
Sad Clown
111
New contributor
Sad Clown is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
When I was your age (15 or so), I also played around with a very similar integral. My take was slightly different.
First of all, for suitable real numbers $s$ (say with positive real part) one actually has the sum
$$sum_-infty^infty binomsk = 2^s.$$
So a more natural integral to consider might be
$$I(s):=int_-infty^infty fracGamma(s+1)Gamma(x+1) Gamma(s+1-x) , dx.$$
You can prove $I(n) = 2^n$ for non-negative integers $n$. The point here is that one can use the reflection formula for the $Gamma$ function. For example, in the easiest case when $s = 0$, the integral becomes
beginalign
& int_-infty^infty frac1Gamma(x+1) Gamma(1-x) , dx \[10pt]
= & int_-infty^infty frac1x Gamma(x) Gamma(1-x) , dx \[10pt]
= & int_-infty^infty fracsin(pi x)x pi , dx = 1.
endalign
For larger integers $n$, you can do pretty much the same thing and then use partial fractions. The same argument will work with your integral, except now the sin integral will go from $0$ to $n$ rather than $infty$, and hence you pick up the corresponding functions.
(and looks like someone has done that).
This doesn't work for general $s$, however. It took me quite a while to work out, but eventually I found a few different arguments to prove that $I(s) = 2^s$, including a fairly clean proof by contour integration.
Of course, all this was in the days before the internet, so I had to figure it out on my own by thinking about it (even though it was surely known before and also not that hard in the end --- there are very few truly deep definite integrals). I feel a little sad that you can just ask the question and someone will come and answer it.
edited 6 mins ago
Michael Hardy
206k23187466
answered 11 mins ago
Sad Clown
111
New contributor
Sad Clown is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 mins ago
Michael Hardy
206k23187466
edited 6 mins ago
Michael Hardy
206k23187466
edited 6 mins ago
Michael Hardy
206k23187466
206k23187466
answered 11 mins ago
Sad Clown
111
New contributor
Sad Clown is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 11 mins ago
Sad Clown
111
answered 11 mins ago
Sad Clown
111
111
New contributor
Sad Clown is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sad Clown is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sad Clown is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thank you so much for the story to go along with your answer. Rest assured I spend much more time playing around with math on my own than I spend asking about it here! I just haven't actually had a formal calculus class so integrals with special functions always tend to scare me off :) But this was too cool to ignore. I'm currently trying to teach myself contour integration, so I'll search for the proof you mention! Once again, thank you.
– TreFox
5 mins ago
add a comment |Â
Thank you so much for the story to go along with your answer. Rest assured I spend much more time playing around with math on my own than I spend asking about it here! I just haven't actually had a formal calculus class so integrals with special functions always tend to scare me off :) But this was too cool to ignore. I'm currently trying to teach myself contour integration, so I'll search for the proof you mention! Once again, thank you.
– TreFox
5 mins ago
Thank you so much for the story to go along with your answer. Rest assured I spend much more time playing around with math on my own than I spend asking about it here! I just haven't actually had a formal calculus class so integrals with special functions always tend to scare me off :) But this was too cool to ignore. I'm currently trying to teach myself contour integration, so I'll search for the proof you mention! Once again, thank you.
– TreFox
5 mins ago
Thank you so much for the story to go along with your answer. Rest assured I spend much more time playing around with math on my own than I spend asking about it here! I just haven't actually had a formal calculus class so integrals with special functions always tend to scare me off :) But this was too cool to ignore. I'm currently trying to teach myself contour integration, so I'll search for the proof you mention! Once again, thank you.
– TreFox
5 mins ago
add a comment |Â
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2
It might be interesting to mention that $$int_0^1 binomx n dx= G_n$$ where $G_n$ are Gregory coefficients.
– Dahaka
2 hours ago
Since $n$ does not change as $k$ goes from $0$ to $n,$ one has $displaystyle I(n) = n!int_0^n fracmathrm dkk!(n-k)!. qquad$
– Michael Hardy
8 mins ago
How did you find it experimentally? Using some tools?
– Dilworth
29 secs ago