How to calculate $int_{0}^ 1 int_{0}^ 1 frac{1}{(1-xy)^ p} dxdy$ with $p>0$.
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This is Hewitt and Stromberg's Real and Abstract Analysis Problem 21.22. My approach was to compare with this question
How to derive $int_0^1 int_0^1 frac{1}{1-xy} ,dy,dx = sum_{n=1}^{infty}frac{1}{n^{2}}$
and use the series expansion of $frac{1}{(1-xy)^ p}$ to calculate this integral. I found this expansion to be
$$frac{1}{(1-xy)^ p}=sum_{n=0}^ infty frac{p cdots (p+n-1)}{n!} (xy)^ n.$$
However, I got this integral is
$$lim_{r to 1} int_0^ r int_0^ r (sum_{n=0}^ infty frac{p cdots (p+n-1)}{n!} (xy)^ n) dxdy = sum_{n=0}^ infty frac{pcdots (p+n-1)}{n!(n+1)^ 2}.$$
But, I don't seem to find if this series converges or diverges, much less what is the value of the integral. Also it says to calculate the following integrals,
$$int_{0}^ 1 int_{0}^ 1 frac{1}{(1-xy)^ p} dydx, int_{0}^ 1 int_{0}^ 1 Bigg|frac{1}{(1-xy)^ p}Bigg| dxdy, int_{0}^ 1 int_{0}^ 1 Bigg|frac{1}{(1-xy)^ p}Bigg| dydx$$
and to compare with Fubini's Theorem.
Have I done something wrong?
Thanks a lot!
integration analysis measure-theory
asked Nov 21 at 21:59
Dora y Diego
93
add a comment |
up vote
0
down vote
favorite
This is Hewitt and Stromberg's Real and Abstract Analysis Problem 21.22. My approach was to compare with this question
How to derive $int_0^1 int_0^1 frac{1}{1-xy} ,dy,dx = sum_{n=1}^{infty}frac{1}{n^{2}}$
and use the series expansion of $frac{1}{(1-xy)^ p}$ to calculate this integral. I found this expansion to be
$$frac{1}{(1-xy)^ p}=sum_{n=0}^ infty frac{p cdots (p+n-1)}{n!} (xy)^ n.$$
However, I got this integral is
$$lim_{r to 1} int_0^ r int_0^ r (sum_{n=0}^ infty frac{p cdots (p+n-1)}{n!} (xy)^ n) dxdy = sum_{n=0}^ infty frac{pcdots (p+n-1)}{n!(n+1)^ 2}.$$
But, I don't seem to find if this series converges or diverges, much less what is the value of the integral. Also it says to calculate the following integrals,
$$int_{0}^ 1 int_{0}^ 1 frac{1}{(1-xy)^ p} dydx, int_{0}^ 1 int_{0}^ 1 Bigg|frac{1}{(1-xy)^ p}Bigg| dxdy, int_{0}^ 1 int_{0}^ 1 Bigg|frac{1}{(1-xy)^ p}Bigg| dydx$$
and to compare with Fubini's Theorem.
Have I done something wrong?
Thanks a lot!
integration analysis measure-theory
asked Nov 21 at 21:59
Dora y Diego
93
You should have $p(p+1)cdots (p+n-1)$, and not $p(p+1)cdots (p+n)$.
– Batominovski
Nov 21 at 22:02
@Batominovski I had it with the $p+n-1$ but I didn't seem to conclude anything so I thought I could just use $p+n$. I've edited now, how can I know if it converges?
– Dora y Diego
Nov 21 at 22:23
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is Hewitt and Stromberg's Real and Abstract Analysis Problem 21.22. My approach was to compare with this question
How to derive $int_0^1 int_0^1 frac{1}{1-xy} ,dy,dx = sum_{n=1}^{infty}frac{1}{n^{2}}$
and use the series expansion of $frac{1}{(1-xy)^ p}$ to calculate this integral. I found this expansion to be
$$frac{1}{(1-xy)^ p}=sum_{n=0}^ infty frac{p cdots (p+n-1)}{n!} (xy)^ n.$$
However, I got this integral is
$$lim_{r to 1} int_0^ r int_0^ r (sum_{n=0}^ infty frac{p cdots (p+n-1)}{n!} (xy)^ n) dxdy = sum_{n=0}^ infty frac{pcdots (p+n-1)}{n!(n+1)^ 2}.$$
But, I don't seem to find if this series converges or diverges, much less what is the value of the integral. Also it says to calculate the following integrals,
$$int_{0}^ 1 int_{0}^ 1 frac{1}{(1-xy)^ p} dydx, int_{0}^ 1 int_{0}^ 1 Bigg|frac{1}{(1-xy)^ p}Bigg| dxdy, int_{0}^ 1 int_{0}^ 1 Bigg|frac{1}{(1-xy)^ p}Bigg| dydx$$
and to compare with Fubini's Theorem.
Have I done something wrong?
Thanks a lot!
integration analysis measure-theory
asked Nov 21 at 21:59
Dora y Diego
93
This is Hewitt and Stromberg's Real and Abstract Analysis Problem 21.22. My approach was to compare with this question
How to derive $int_0^1 int_0^1 frac{1}{1-xy} ,dy,dx = sum_{n=1}^{infty}frac{1}{n^{2}}$
and use the series expansion of $frac{1}{(1-xy)^ p}$ to calculate this integral. I found this expansion to be
$$frac{1}{(1-xy)^ p}=sum_{n=0}^ infty frac{p cdots (p+n-1)}{n!} (xy)^ n.$$
However, I got this integral is
$$lim_{r to 1} int_0^ r int_0^ r (sum_{n=0}^ infty frac{p cdots (p+n-1)}{n!} (xy)^ n) dxdy = sum_{n=0}^ infty frac{pcdots (p+n-1)}{n!(n+1)^ 2}.$$
But, I don't seem to find if this series converges or diverges, much less what is the value of the integral. Also it says to calculate the following integrals,
$$int_{0}^ 1 int_{0}^ 1 frac{1}{(1-xy)^ p} dydx, int_{0}^ 1 int_{0}^ 1 Bigg|frac{1}{(1-xy)^ p}Bigg| dxdy, int_{0}^ 1 int_{0}^ 1 Bigg|frac{1}{(1-xy)^ p}Bigg| dydx$$
and to compare with Fubini's Theorem.
Have I done something wrong?
Thanks a lot!
integration analysis measure-theory
integration analysis measure-theory
asked Nov 21 at 21:59
Dora y Diego
93
asked Nov 21 at 21:59
Dora y Diego
93
edited Nov 21 at 22:20
asked Nov 21 at 21:59
Dora y Diego
93
asked Nov 21 at 21:59
Dora y Diego
93
asked Nov 21 at 21:59
Dora y Diego
93
93
You should have $p(p+1)cdots (p+n-1)$, and not $p(p+1)cdots (p+n)$.
– Batominovski
Nov 21 at 22:02
@Batominovski I had it with the $p+n-1$ but I didn't seem to conclude anything so I thought I could just use $p+n$. I've edited now, how can I know if it converges?
– Dora y Diego
Nov 21 at 22:23
add a comment |
You should have $p(p+1)cdots (p+n-1)$, and not $p(p+1)cdots (p+n)$.
– Batominovski
Nov 21 at 22:02
@Batominovski I had it with the $p+n-1$ but I didn't seem to conclude anything so I thought I could just use $p+n$. I've edited now, how can I know if it converges?
– Dora y Diego
Nov 21 at 22:23
You should have $p(p+1)cdots (p+n-1)$, and not $p(p+1)cdots (p+n)$.
– Batominovski
Nov 21 at 22:02
You should have $p(p+1)cdots (p+n-1)$, and not $p(p+1)cdots (p+n)$.
– Batominovski
Nov 21 at 22:02
@Batominovski I had it with the $p+n-1$ but I didn't seem to conclude anything so I thought I could just use $p+n$. I've edited now, how can I know if it converges?
– Dora y Diego
Nov 21 at 22:23
@Batominovski I had it with the $p+n-1$ but I didn't seem to conclude anything so I thought I could just use $p+n$. I've edited now, how can I know if it converges?
– Dora y Diego
Nov 21 at 22:23
add a comment |
1 Answer
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This is a well-known integral for $0<p<1$ and involves the Harmonic number:
$$frac{H_{1-p}}{1-p}$$
answered Nov 21 at 22:03
David G. Stork
9,32721232
I do not see why you had a downvote. $to +1$.
– Claude Leibovici
Nov 22 at 6:53
@ClaudeLeibovici: Nor did I. Thanks.
– David G. Stork
Nov 22 at 7:15
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is a well-known integral for $0<p<1$ and involves the Harmonic number:
$$frac{H_{1-p}}{1-p}$$
answered Nov 21 at 22:03
David G. Stork
9,32721232
I do not see why you had a downvote. $to +1$.
– Claude Leibovici
Nov 22 at 6:53
@ClaudeLeibovici: Nor did I. Thanks.
– David G. Stork
Nov 22 at 7:15
add a comment |
up vote
1
down vote
This is a well-known integral for $0<p<1$ and involves the Harmonic number:
$$frac{H_{1-p}}{1-p}$$
answered Nov 21 at 22:03
David G. Stork
9,32721232
I do not see why you had a downvote. $to +1$.
– Claude Leibovici
Nov 22 at 6:53
@ClaudeLeibovici: Nor did I. Thanks.
– David G. Stork
Nov 22 at 7:15
add a comment |
up vote
1
down vote
up vote
1
down vote
This is a well-known integral for $0<p<1$ and involves the Harmonic number:
$$frac{H_{1-p}}{1-p}$$
answered Nov 21 at 22:03
David G. Stork
9,32721232
This is a well-known integral for $0<p<1$ and involves the Harmonic number:
$$frac{H_{1-p}}{1-p}$$
answered Nov 21 at 22:03
David G. Stork
9,32721232
answered Nov 21 at 22:03
David G. Stork
9,32721232
answered Nov 21 at 22:03
David G. Stork
9,32721232
answered Nov 21 at 22:03
David G. Stork
9,32721232
9,32721232
I do not see why you had a downvote. $to +1$.
– Claude Leibovici
Nov 22 at 6:53
@ClaudeLeibovici: Nor did I. Thanks.
– David G. Stork
Nov 22 at 7:15
add a comment |
I do not see why you had a downvote. $to +1$.
– Claude Leibovici
Nov 22 at 6:53
@ClaudeLeibovici: Nor did I. Thanks.
– David G. Stork
Nov 22 at 7:15
I do not see why you had a downvote. $to +1$.
– Claude Leibovici
Nov 22 at 6:53
I do not see why you had a downvote. $to +1$.
– Claude Leibovici
Nov 22 at 6:53
@ClaudeLeibovici: Nor did I. Thanks.
– David G. Stork
Nov 22 at 7:15
@ClaudeLeibovici: Nor did I. Thanks.
– David G. Stork
Nov 22 at 7:15
add a comment |
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You should have $p(p+1)cdots (p+n-1)$, and not $p(p+1)cdots (p+n)$.
– Batominovski
Nov 21 at 22:02
@Batominovski I had it with the $p+n-1$ but I didn't seem to conclude anything so I thought I could just use $p+n$. I've edited now, how can I know if it converges?
– Dora y Diego
Nov 21 at 22:23