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Evaluate $int_0^1 frac11+x^mdx$ for $m>0 $
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I came across the sum:
$$I(m)=sum_n=0^infty frac(-1)^nmn+1 $$
where $m>0$.
It's easy to see that this sum is equal to:
$$int_0^1 frac11+x^mdx$$
for $m>0$
So I tried my hardest to find a way of evaluating it. First I thought using Euler's Beta Function but to no avail. Then I decided to brute force a partial fraction decomposition, and I got this ugly mess (which is more than like incorrect):
$$I(m)=sum_n=0^infty frac(-1)^nmn+1 = sum_k=0^infty A_kln|1-e^-(i(2k+1))/m| $$Where $A_k$ equals
$$A_k= frac1prod_n=0^k(e^i(2k+1)-e^ipi(2n+1)/m)prod_n=k^infty(e^ikpi/m-e^i(2n+3)) $$
I would check my work, but it's really tricky to check so I'm just hoping it's right. Is there a way of evaluating the integral that's less ugly than what I did?
integration summation
asked Jul 18 at 3:05
Tom Himler
654213
add a comment |Â
up vote
6
down vote
favorite
I came across the sum:
$$I(m)=sum_n=0^infty frac(-1)^nmn+1 $$
where $m>0$.
It's easy to see that this sum is equal to:
$$int_0^1 frac11+x^mdx$$
for $m>0$
So I tried my hardest to find a way of evaluating it. First I thought using Euler's Beta Function but to no avail. Then I decided to brute force a partial fraction decomposition, and I got this ugly mess (which is more than like incorrect):
$$I(m)=sum_n=0^infty frac(-1)^nmn+1 = sum_k=0^infty A_kln|1-e^-(i(2k+1))/m| $$Where $A_k$ equals
$$A_k= frac1prod_n=0^k(e^i(2k+1)-e^ipi(2n+1)/m)prod_n=k^infty(e^ikpi/m-e^i(2n+3)) $$
I would check my work, but it's really tricky to check so I'm just hoping it's right. Is there a way of evaluating the integral that's less ugly than what I did?
integration summation
asked Jul 18 at 3:05
Tom Himler
654213
After looking at the $A_k$ I'm sure it's wrong. I wanted when k=0 for the first infinite product to be equal to 1. Oh well.
– Tom Himler
Jul 18 at 3:06
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I came across the sum:
$$I(m)=sum_n=0^infty frac(-1)^nmn+1 $$
where $m>0$.
It's easy to see that this sum is equal to:
$$int_0^1 frac11+x^mdx$$
for $m>0$
So I tried my hardest to find a way of evaluating it. First I thought using Euler's Beta Function but to no avail. Then I decided to brute force a partial fraction decomposition, and I got this ugly mess (which is more than like incorrect):
$$I(m)=sum_n=0^infty frac(-1)^nmn+1 = sum_k=0^infty A_kln|1-e^-(i(2k+1))/m| $$Where $A_k$ equals
$$A_k= frac1prod_n=0^k(e^i(2k+1)-e^ipi(2n+1)/m)prod_n=k^infty(e^ikpi/m-e^i(2n+3)) $$
I would check my work, but it's really tricky to check so I'm just hoping it's right. Is there a way of evaluating the integral that's less ugly than what I did?
integration summation
asked Jul 18 at 3:05
Tom Himler
654213
I came across the sum:
$$I(m)=sum_n=0^infty frac(-1)^nmn+1 $$
where $m>0$.
It's easy to see that this sum is equal to:
$$int_0^1 frac11+x^mdx$$
for $m>0$
So I tried my hardest to find a way of evaluating it. First I thought using Euler's Beta Function but to no avail. Then I decided to brute force a partial fraction decomposition, and I got this ugly mess (which is more than like incorrect):
$$I(m)=sum_n=0^infty frac(-1)^nmn+1 = sum_k=0^infty A_kln|1-e^-(i(2k+1))/m| $$Where $A_k$ equals
$$A_k= frac1prod_n=0^k(e^i(2k+1)-e^ipi(2n+1)/m)prod_n=k^infty(e^ikpi/m-e^i(2n+3)) $$
I would check my work, but it's really tricky to check so I'm just hoping it's right. Is there a way of evaluating the integral that's less ugly than what I did?
integration summation
asked Jul 18 at 3:05
Tom Himler
654213
asked Jul 18 at 3:05
Tom Himler
654213
asked Jul 18 at 3:05
Tom Himler
654213
asked Jul 18 at 3:05
Tom Himler
654213
654213
After looking at the $A_k$ I'm sure it's wrong. I wanted when k=0 for the first infinite product to be equal to 1. Oh well.
– Tom Himler
Jul 18 at 3:06
add a comment |Â
After looking at the $A_k$ I'm sure it's wrong. I wanted when k=0 for the first infinite product to be equal to 1. Oh well.
– Tom Himler
Jul 18 at 3:06
After looking at the $A_k$ I'm sure it's wrong. I wanted when k=0 for the first infinite product to be equal to 1. Oh well.
– Tom Himler
Jul 18 at 3:06
After looking at the $A_k$ I'm sure it's wrong. I wanted when k=0 for the first infinite product to be equal to 1. Oh well.
– Tom Himler
Jul 18 at 3:06
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
Using the digamma function $$int_0^1 frac11+x^mdx=fracpsi ^(0)left(fracm+12 mright)-psi ^(0)left(frac12
mright)2 m$$
Using the Hurwitz-Lerch transcendent function
$$int_0^1 frac11+x^mdx=fracPhi left(-1,1,frac1mright)m$$
answered Jul 18 at 4:27
Claude Leibovici
112k1055126
That’s good enough for me, thank you.
– Tom Himler
Jul 18 at 11:29
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Using the digamma function $$int_0^1 frac11+x^mdx=fracpsi ^(0)left(fracm+12 mright)-psi ^(0)left(frac12
mright)2 m$$
Using the Hurwitz-Lerch transcendent function
$$int_0^1 frac11+x^mdx=fracPhi left(-1,1,frac1mright)m$$
answered Jul 18 at 4:27
Claude Leibovici
112k1055126
That’s good enough for me, thank you.
– Tom Himler
Jul 18 at 11:29
add a comment |Â
up vote
5
down vote
accepted
Using the digamma function $$int_0^1 frac11+x^mdx=fracpsi ^(0)left(fracm+12 mright)-psi ^(0)left(frac12
mright)2 m$$
Using the Hurwitz-Lerch transcendent function
$$int_0^1 frac11+x^mdx=fracPhi left(-1,1,frac1mright)m$$
answered Jul 18 at 4:27
Claude Leibovici
112k1055126
That’s good enough for me, thank you.
– Tom Himler
Jul 18 at 11:29
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Using the digamma function $$int_0^1 frac11+x^mdx=fracpsi ^(0)left(fracm+12 mright)-psi ^(0)left(frac12
mright)2 m$$
Using the Hurwitz-Lerch transcendent function
$$int_0^1 frac11+x^mdx=fracPhi left(-1,1,frac1mright)m$$
answered Jul 18 at 4:27
Claude Leibovici
112k1055126
Using the digamma function $$int_0^1 frac11+x^mdx=fracpsi ^(0)left(fracm+12 mright)-psi ^(0)left(frac12
mright)2 m$$
Using the Hurwitz-Lerch transcendent function
$$int_0^1 frac11+x^mdx=fracPhi left(-1,1,frac1mright)m$$
answered Jul 18 at 4:27
Claude Leibovici
112k1055126
answered Jul 18 at 4:27
Claude Leibovici
112k1055126
answered Jul 18 at 4:27
Claude Leibovici
112k1055126
answered Jul 18 at 4:27
Claude Leibovici
112k1055126
112k1055126
That’s good enough for me, thank you.
– Tom Himler
Jul 18 at 11:29
add a comment |Â
That’s good enough for me, thank you.
– Tom Himler
Jul 18 at 11:29
That’s good enough for me, thank you.
– Tom Himler
Jul 18 at 11:29
That’s good enough for me, thank you.
– Tom Himler
Jul 18 at 11:29
add a comment |Â
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After looking at the $A_k$ I'm sure it's wrong. I wanted when k=0 for the first infinite product to be equal to 1. Oh well.
– Tom Himler
Jul 18 at 3:06