Mixing
Double summation with improper integral
Clash Royale CLAN TAG#URR8PPP
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So my friend sent me this really interesting problem. It goes:
Evaluate the following expression:
$$ sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx $$
Here is my approach:
First evaluate the integral:
$$ frac1b! int_0^infty fracx^be^ax $$
This can be done using integration by parts and we get:
$$ frac1b! fracba int_0^infty fracx^b-1e^ax$$
We can do this $ b $ times until we get:
$$ frac1b! frac(b)(b-a).....(b-b+1)a^b int_0^infty fracx^b-be^ax $$
and hence we end up with:
$$ frac1b! fracb!a^b (frac-1 e^-axaBig|_0^infty) = frac1a^b+1$$
Now we can apply the sum of GP to infinity formula and we get:
$$ sum_a=2^infty sum_b=1^infty frac1a^b+1 = sum_a=2^infty fracfrac1a^21-frac1a $$
This is a telescoping series and we end up with $$ frac1a-1 = frac12-1 = 1 $$
Do you guys have any other ways of solving this problem? Please do share it here.
integration summation geometric-progressions
asked 1 hour ago
Tusky
1546
add a comment |Â
up vote
3
down vote
favorite
So my friend sent me this really interesting problem. It goes:
Evaluate the following expression:
$$ sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx $$
Here is my approach:
First evaluate the integral:
$$ frac1b! int_0^infty fracx^be^ax $$
This can be done using integration by parts and we get:
$$ frac1b! fracba int_0^infty fracx^b-1e^ax$$
We can do this $ b $ times until we get:
$$ frac1b! frac(b)(b-a).....(b-b+1)a^b int_0^infty fracx^b-be^ax $$
and hence we end up with:
$$ frac1b! fracb!a^b (frac-1 e^-axaBig|_0^infty) = frac1a^b+1$$
Now we can apply the sum of GP to infinity formula and we get:
$$ sum_a=2^infty sum_b=1^infty frac1a^b+1 = sum_a=2^infty fracfrac1a^21-frac1a $$
This is a telescoping series and we end up with $$ frac1a-1 = frac12-1 = 1 $$
Do you guys have any other ways of solving this problem? Please do share it here.
integration summation geometric-progressions
asked 1 hour ago
Tusky
1546
I think this is the simplest way have you done!
– Nosrati
1 hour ago
One approach: interchange summations and integration (allowed by Tonelli's theorem) and perform the sum's ($e^x$ and geometrical series) to end up with a simple integral.
– Winther
57 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
So my friend sent me this really interesting problem. It goes:
Evaluate the following expression:
$$ sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx $$
Here is my approach:
First evaluate the integral:
$$ frac1b! int_0^infty fracx^be^ax $$
This can be done using integration by parts and we get:
$$ frac1b! fracba int_0^infty fracx^b-1e^ax$$
We can do this $ b $ times until we get:
$$ frac1b! frac(b)(b-a).....(b-b+1)a^b int_0^infty fracx^b-be^ax $$
and hence we end up with:
$$ frac1b! fracb!a^b (frac-1 e^-axaBig|_0^infty) = frac1a^b+1$$
Now we can apply the sum of GP to infinity formula and we get:
$$ sum_a=2^infty sum_b=1^infty frac1a^b+1 = sum_a=2^infty fracfrac1a^21-frac1a $$
This is a telescoping series and we end up with $$ frac1a-1 = frac12-1 = 1 $$
Do you guys have any other ways of solving this problem? Please do share it here.
integration summation geometric-progressions
asked 1 hour ago
Tusky
1546
So my friend sent me this really interesting problem. It goes:
Evaluate the following expression:
$$ sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx $$
Here is my approach:
First evaluate the integral:
$$ frac1b! int_0^infty fracx^be^ax $$
This can be done using integration by parts and we get:
$$ frac1b! fracba int_0^infty fracx^b-1e^ax$$
We can do this $ b $ times until we get:
$$ frac1b! frac(b)(b-a).....(b-b+1)a^b int_0^infty fracx^b-be^ax $$
and hence we end up with:
$$ frac1b! fracb!a^b (frac-1 e^-axaBig|_0^infty) = frac1a^b+1$$
Now we can apply the sum of GP to infinity formula and we get:
$$ sum_a=2^infty sum_b=1^infty frac1a^b+1 = sum_a=2^infty fracfrac1a^21-frac1a $$
This is a telescoping series and we end up with $$ frac1a-1 = frac12-1 = 1 $$
Do you guys have any other ways of solving this problem? Please do share it here.
integration summation geometric-progressions
integration summation geometric-progressions
asked 1 hour ago
Tusky
1546
asked 1 hour ago
Tusky
1546
asked 1 hour ago
Tusky
1546
asked 1 hour ago
Tusky
1546
asked 1 hour ago
Tusky
1546
1546
I think this is the simplest way have you done!
– Nosrati
1 hour ago
One approach: interchange summations and integration (allowed by Tonelli's theorem) and perform the sum's ($e^x$ and geometrical series) to end up with a simple integral.
– Winther
57 mins ago
add a comment |Â
I think this is the simplest way have you done!
– Nosrati
1 hour ago
One approach: interchange summations and integration (allowed by Tonelli's theorem) and perform the sum's ($e^x$ and geometrical series) to end up with a simple integral.
– Winther
57 mins ago
I think this is the simplest way have you done!
– Nosrati
1 hour ago
I think this is the simplest way have you done!
– Nosrati
1 hour ago
One approach: interchange summations and integration (allowed by Tonelli's theorem) and perform the sum's ($e^x$ and geometrical series) to end up with a simple integral.
– Winther
57 mins ago
One approach: interchange summations and integration (allowed by Tonelli's theorem) and perform the sum's ($e^x$ and geometrical series) to end up with a simple integral.
– Winther
57 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
since $fracx^be^ax b!$ is non-negative, Tonelli's theorem for iterated integrals/sums allows us to interchange integrals and sums without worry. Then:
beginalign
&sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx \
&=int_0^infty sum_a=2^infty e^-ax sum_b=1^infty fracx^b b! dx \
&= int_0^infty underbraceleft(sum_a=2^infty (e^-x)^aright)_textgeometric series overbraceleft(sum_b=0^infty fracx^b b!-1right)^textseries definition of $e^x$ dx \
&= int_0^infty frac1e^x(e^x-1)(e^x-1)dx \
&= int_0^infty e^-x dx \&= 1.endalign
answered 54 mins ago
Calvin Khor
9,17521233
add a comment |Â
up vote
2
down vote
For the first part I often use the Laplace transform:
$$frac1b! int_0^infty fracx^be^ax dx = frac1b! int_0^infty x^be^-ax dx = frac1b! cal L(x^b)Big|_s=a = frac1b! fracb!a^b+1 = frac1a^b+1$$
this make it easier.
answered 1 hour ago
Nosrati
23.3k61952
And how do you evaluate the Laplace transform? I don't see how it's easier unless you 'cheat' and look up the result in a table.
– Winther
1 hour ago
I don't think so, you are kidding $$cal L(x^b)=dfracGamma(b+1)s^b+1$$
– Nosrati
1 hour ago
You are not answering my question.
– Winther
59 mins ago
Laplace transform is a tool. We use it's formulas whatever we want!
– Nosrati
54 mins ago
add a comment |Â
up vote
1
down vote
The integral is of the Gamma type,
$$int_0^infty fracx^be^ax dx=frac1a^b+1int_0^infty t^be^-t dx =fracb!a^b+1.$$
Then
$$sum_a=2^infty sum_b=1^inftyfrac1a^b+1=sum_a=2^infty frac1a^2left(1-dfrac1aright)=sum_a=2^infty frac1a(a-1)$$ is indeed a telescoping sum, giving
$$frac12-1.$$
answered 27 mins ago
Yves Daoust
116k667211
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
since $fracx^be^ax b!$ is non-negative, Tonelli's theorem for iterated integrals/sums allows us to interchange integrals and sums without worry. Then:
beginalign
&sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx \
&=int_0^infty sum_a=2^infty e^-ax sum_b=1^infty fracx^b b! dx \
&= int_0^infty underbraceleft(sum_a=2^infty (e^-x)^aright)_textgeometric series overbraceleft(sum_b=0^infty fracx^b b!-1right)^textseries definition of $e^x$ dx \
&= int_0^infty frac1e^x(e^x-1)(e^x-1)dx \
&= int_0^infty e^-x dx \&= 1.endalign
answered 54 mins ago
Calvin Khor
9,17521233
add a comment |Â
up vote
3
down vote
since $fracx^be^ax b!$ is non-negative, Tonelli's theorem for iterated integrals/sums allows us to interchange integrals and sums without worry. Then:
beginalign
&sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx \
&=int_0^infty sum_a=2^infty e^-ax sum_b=1^infty fracx^b b! dx \
&= int_0^infty underbraceleft(sum_a=2^infty (e^-x)^aright)_textgeometric series overbraceleft(sum_b=0^infty fracx^b b!-1right)^textseries definition of $e^x$ dx \
&= int_0^infty frac1e^x(e^x-1)(e^x-1)dx \
&= int_0^infty e^-x dx \&= 1.endalign
answered 54 mins ago
Calvin Khor
9,17521233
add a comment |Â
up vote
3
down vote
up vote
3
down vote
since $fracx^be^ax b!$ is non-negative, Tonelli's theorem for iterated integrals/sums allows us to interchange integrals and sums without worry. Then:
beginalign
&sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx \
&=int_0^infty sum_a=2^infty e^-ax sum_b=1^infty fracx^b b! dx \
&= int_0^infty underbraceleft(sum_a=2^infty (e^-x)^aright)_textgeometric series overbraceleft(sum_b=0^infty fracx^b b!-1right)^textseries definition of $e^x$ dx \
&= int_0^infty frac1e^x(e^x-1)(e^x-1)dx \
&= int_0^infty e^-x dx \&= 1.endalign
answered 54 mins ago
Calvin Khor
9,17521233
since $fracx^be^ax b!$ is non-negative, Tonelli's theorem for iterated integrals/sums allows us to interchange integrals and sums without worry. Then:
beginalign
&sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx \
&=int_0^infty sum_a=2^infty e^-ax sum_b=1^infty fracx^b b! dx \
&= int_0^infty underbraceleft(sum_a=2^infty (e^-x)^aright)_textgeometric series overbraceleft(sum_b=0^infty fracx^b b!-1right)^textseries definition of $e^x$ dx \
&= int_0^infty frac1e^x(e^x-1)(e^x-1)dx \
&= int_0^infty e^-x dx \&= 1.endalign
answered 54 mins ago
Calvin Khor
9,17521233
edited 47 mins ago
answered 54 mins ago
Calvin Khor
9,17521233
answered 54 mins ago
Calvin Khor
9,17521233
answered 54 mins ago
Calvin Khor
9,17521233
9,17521233
add a comment |Â
add a comment |Â
up vote
2
down vote
For the first part I often use the Laplace transform:
$$frac1b! int_0^infty fracx^be^ax dx = frac1b! int_0^infty x^be^-ax dx = frac1b! cal L(x^b)Big|_s=a = frac1b! fracb!a^b+1 = frac1a^b+1$$
this make it easier.
answered 1 hour ago
Nosrati
23.3k61952
And how do you evaluate the Laplace transform? I don't see how it's easier unless you 'cheat' and look up the result in a table.
– Winther
1 hour ago
I don't think so, you are kidding $$cal L(x^b)=dfracGamma(b+1)s^b+1$$
– Nosrati
1 hour ago
You are not answering my question.
– Winther
59 mins ago
Laplace transform is a tool. We use it's formulas whatever we want!
– Nosrati
54 mins ago
add a comment |Â
up vote
2
down vote
For the first part I often use the Laplace transform:
$$frac1b! int_0^infty fracx^be^ax dx = frac1b! int_0^infty x^be^-ax dx = frac1b! cal L(x^b)Big|_s=a = frac1b! fracb!a^b+1 = frac1a^b+1$$
this make it easier.
answered 1 hour ago
Nosrati
23.3k61952
And how do you evaluate the Laplace transform? I don't see how it's easier unless you 'cheat' and look up the result in a table.
– Winther
1 hour ago
I don't think so, you are kidding $$cal L(x^b)=dfracGamma(b+1)s^b+1$$
– Nosrati
1 hour ago
You are not answering my question.
– Winther
59 mins ago
Laplace transform is a tool. We use it's formulas whatever we want!
– Nosrati
54 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For the first part I often use the Laplace transform:
$$frac1b! int_0^infty fracx^be^ax dx = frac1b! int_0^infty x^be^-ax dx = frac1b! cal L(x^b)Big|_s=a = frac1b! fracb!a^b+1 = frac1a^b+1$$
this make it easier.
answered 1 hour ago
Nosrati
23.3k61952
For the first part I often use the Laplace transform:
$$frac1b! int_0^infty fracx^be^ax dx = frac1b! int_0^infty x^be^-ax dx = frac1b! cal L(x^b)Big|_s=a = frac1b! fracb!a^b+1 = frac1a^b+1$$
this make it easier.
answered 1 hour ago
Nosrati
23.3k61952
answered 1 hour ago
Nosrati
23.3k61952
answered 1 hour ago
Nosrati
23.3k61952
answered 1 hour ago
Nosrati
23.3k61952
23.3k61952
And how do you evaluate the Laplace transform? I don't see how it's easier unless you 'cheat' and look up the result in a table.
– Winther
1 hour ago
I don't think so, you are kidding $$cal L(x^b)=dfracGamma(b+1)s^b+1$$
– Nosrati
1 hour ago
You are not answering my question.
– Winther
59 mins ago
Laplace transform is a tool. We use it's formulas whatever we want!
– Nosrati
54 mins ago
add a comment |Â
And how do you evaluate the Laplace transform? I don't see how it's easier unless you 'cheat' and look up the result in a table.
– Winther
1 hour ago
I don't think so, you are kidding $$cal L(x^b)=dfracGamma(b+1)s^b+1$$
– Nosrati
1 hour ago
You are not answering my question.
– Winther
59 mins ago
Laplace transform is a tool. We use it's formulas whatever we want!
– Nosrati
54 mins ago
And how do you evaluate the Laplace transform? I don't see how it's easier unless you 'cheat' and look up the result in a table.
– Winther
1 hour ago
And how do you evaluate the Laplace transform? I don't see how it's easier unless you 'cheat' and look up the result in a table.
– Winther
1 hour ago
I don't think so, you are kidding $$cal L(x^b)=dfracGamma(b+1)s^b+1$$
– Nosrati
1 hour ago
I don't think so, you are kidding $$cal L(x^b)=dfracGamma(b+1)s^b+1$$
– Nosrati
1 hour ago
You are not answering my question.
– Winther
59 mins ago
You are not answering my question.
– Winther
59 mins ago
Laplace transform is a tool. We use it's formulas whatever we want!
– Nosrati
54 mins ago
Laplace transform is a tool. We use it's formulas whatever we want!
– Nosrati
54 mins ago
add a comment |Â
up vote
1
down vote
The integral is of the Gamma type,
$$int_0^infty fracx^be^ax dx=frac1a^b+1int_0^infty t^be^-t dx =fracb!a^b+1.$$
Then
$$sum_a=2^infty sum_b=1^inftyfrac1a^b+1=sum_a=2^infty frac1a^2left(1-dfrac1aright)=sum_a=2^infty frac1a(a-1)$$ is indeed a telescoping sum, giving
$$frac12-1.$$
answered 27 mins ago
Yves Daoust
116k667211
add a comment |Â
up vote
1
down vote
The integral is of the Gamma type,
$$int_0^infty fracx^be^ax dx=frac1a^b+1int_0^infty t^be^-t dx =fracb!a^b+1.$$
Then
$$sum_a=2^infty sum_b=1^inftyfrac1a^b+1=sum_a=2^infty frac1a^2left(1-dfrac1aright)=sum_a=2^infty frac1a(a-1)$$ is indeed a telescoping sum, giving
$$frac12-1.$$
answered 27 mins ago
Yves Daoust
116k667211
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The integral is of the Gamma type,
$$int_0^infty fracx^be^ax dx=frac1a^b+1int_0^infty t^be^-t dx =fracb!a^b+1.$$
Then
$$sum_a=2^infty sum_b=1^inftyfrac1a^b+1=sum_a=2^infty frac1a^2left(1-dfrac1aright)=sum_a=2^infty frac1a(a-1)$$ is indeed a telescoping sum, giving
$$frac12-1.$$
answered 27 mins ago
Yves Daoust
116k667211
The integral is of the Gamma type,
$$int_0^infty fracx^be^ax dx=frac1a^b+1int_0^infty t^be^-t dx =fracb!a^b+1.$$
Then
$$sum_a=2^infty sum_b=1^inftyfrac1a^b+1=sum_a=2^infty frac1a^2left(1-dfrac1aright)=sum_a=2^infty frac1a(a-1)$$ is indeed a telescoping sum, giving
$$frac12-1.$$
answered 27 mins ago
Yves Daoust
116k667211
answered 27 mins ago
Yves Daoust
116k667211
answered 27 mins ago
Yves Daoust
116k667211
answered 27 mins ago
Yves Daoust
116k667211
116k667211
add a comment |Â
add a comment |Â
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I think this is the simplest way have you done!
– Nosrati
1 hour ago
One approach: interchange summations and integration (allowed by Tonelli's theorem) and perform the sum's ($e^x$ and geometrical series) to end up with a simple integral.
– Winther
57 mins ago