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!










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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

















up vote
0
down vote

favorite
1












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!










share|cite|improve this question
























  • 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















up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





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!










share|cite|improve this question















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






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edited Nov 21 at 22:20

























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




















  • 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












1 Answer
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This is a well-known integral for $0<p<1$ and involves the Harmonic number:



$$frac{H_{1-p}}{1-p}$$






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  • 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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes








up vote
1
down vote













This is a well-known integral for $0<p<1$ and involves the Harmonic number:



$$frac{H_{1-p}}{1-p}$$






share|cite|improve this answer





















  • 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















up vote
1
down vote













This is a well-known integral for $0<p<1$ and involves the Harmonic number:



$$frac{H_{1-p}}{1-p}$$






share|cite|improve this answer





















  • 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













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:



$$frac{H_{1-p}}{1-p}$$






share|cite|improve this answer












This is a well-known integral for $0<p<1$ and involves the Harmonic number:



$$frac{H_{1-p}}{1-p}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 at 22:03









David G. Stork

9,12421232




9,12421232












  • 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










  • @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


















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