Logarithm,Double integrals

Calculus Level 5

∫ 0 1 ∫ 0 1 ln ⁡ 2 ( 1 − x y ) x y d x d y \displaystyle \int _{ 0 }^{ 1 }\int _{ 0 }^{ 1 } \dfrac { { \ln }^{ 2 }(1-xy) }{ xy } \ dx \ dy

If the double integral above can be represented as π A B \dfrac { { \pi }^{ A } }{ B } , where A , B A,B are positive integers, find A 2 + B 2 − 32236 A^2+B^2-32236 .

This is original. Check out my Set


The answer is 180.

Ronak Agarwal by Ronak Agarwal , 23, India

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3 solutions

Pratik Shastri
Mar 6, 2015

I = ∫ 0 1 ∫ 0 1 log ⁡ 2 ( 1 − x y ) x y d x d y = ∫ 0 1 ∫ 0 1 log ⁡ ( 1 − x y ) log ⁡ ( 1 − x y ) x y d x d y = ∫ 0 1 ∫ 0 1 ∑ m = 1 ∞ ∑ n = 1 ∞ 1 x y ( x y ) m + n m n d x d y = ∫ 0 1 ∫ 0 1 ∑ m = 1 ∞ ∑ n = 1 ∞ ( x y ) m + n − 1 m n d x d y = ∑ m = 1 ∞ ∑ n = 1 ∞ 1 ( m + n ) 2 m n = T ( 1 , 1 , 2 ) = ζ ( 4 ) 2 \begin{aligned} I&=\int_{0}^{1} \int_{0}^{1} \dfrac{\log^2(1-xy)}{xy} \mathrm{d}x \ \mathrm{d}y\\ &=\int_{0}^{1} \int_{0}^{1} \dfrac{\log(1-xy) \log(1-xy)}{xy} \mathrm{d}x \ \mathrm{d}y\\ &= \int_{0}^{1} \int_{0}^{1} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{xy} \dfrac{(xy)^{m+n}}{mn} \mathrm{d}x \ \mathrm{d}y\\ &= \int_{0}^{1} \int_{0}^{1} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{(xy)^{m+n-1}}{mn} \mathrm{d}x \ \mathrm{d}y\\ &=\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{(m+n)^2mn}\\ &=T(1,1,2)\\ &=\dfrac{\zeta (4)}{2} \end{aligned}

For the double last sum, refer to the following three papers -

1

2

3

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So do you know the identity from the begining that ∑ m = 0 ∞ ∑ n = 0 ∞ 1 ( m + n ) 2 m n = ζ ( 4 ) 2 \displaystyle \sum_{m=0}^{\infty} \displaystyle \sum_{n=0}^{\infty} \dfrac{1}{ (m+n)^2mn } = \dfrac{\zeta(4)}{2} , because this was the main part of the problem.

Ronak Agarwal - 6 years, 3 months ago

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I wasn't very sure about evaluating the double sum. I referred to the following two papers -

1

2

Pratik Shastri - 6 years, 3 months ago

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Well @Azhaghu Roopesh M has solved this problem too, I wonder how he has evaluated this sum. Can you please post the solution about the summation Azhaghu. Thanks, well I believe you are practicing double summations it seems.

Ronak Agarwal - 6 years, 3 months ago

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@Ronak Agarwal – The initial steps are too simple , just as Pratik has done it . Now for the double summation what I did was ,

∑ m = 0 ∞ ∑ n = 0 ∞ 1 ( m + n ) 2 m n = ∑ m = 0 ∞ ∑ n = 0 ∞ 1 2 ( 2 m n m 2 n 2 ( m + n ) 2 ) = ∑ m = 0 ∞ ∑ n = 0 ∞ 1 2 ( ( m + n ) 2 − m 2 − n 2 m 2 n 2 ( m + n ) 2 ) = 1 2 ( ∑ m = 0 ∞ ∑ n = 0 ∞ 1 m 2 n 2 − ∑ m = 0 ∞ ∑ n = 0 ∞ 1 n 2 ( m + n ) 2 − ∑ m = 0 ∞ ∑ n = 0 ∞ 1 m 2 ( m + n ) 2 ) = 1 2 ( π 4 36 − 2 × π 4 120 ) = π 4 180 \displaystyle \sum_{m=0}^{\infty} \displaystyle \sum_{n=0}^{\infty} \dfrac{1}{ (m+n)^2mn } \\= \displaystyle \sum_{m=0}^{\infty} \displaystyle \sum_{n=0}^{\infty} \frac{1}{2} \left ( \dfrac{2mn}{m^{2}n^{2}(m+n)^{2}} \right ) \\= \displaystyle \sum_{m=0}^{\infty} \displaystyle \sum_{n=0}^{\infty} \frac{1}{2} \left ( \dfrac{(m+n)^{2}-m^{2}-n^{2}}{m^{2}n^{2}(m+n)^{2}} \right ) \\= \dfrac{1}{2} \left ( \displaystyle \sum_{m=0}^{\infty} \displaystyle \sum_{n=0}^{\infty} \dfrac{1}{m^{2}n^{2}} - \displaystyle \sum_{m=0}^{\infty} \displaystyle \sum_{n=0}^{\infty} \dfrac{1}{n^{2}(m+n)^{2}} - \displaystyle \sum_{m=0}^{\infty} \displaystyle \sum_{n=0}^{\infty} \dfrac{1}{m^{2}(m+n)^{2}} \right ) \\= \frac{1}{2} \left ( \dfrac{\pi^{4}}{36} - 2\times \dfrac{\pi^{4}}{120} \right ) \\= \dfrac{\pi^{4}}{180} .

A Former Brilliant Member - 6 years, 3 months ago

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@A Former Brilliant Member – Kindly enlighten me how you evaluated ∑ n = 1 ∞ ∑ m = 1 ∞ 1 m 2 ( m + n ) 2 \displaystyle \sum _{ n=1 }^{ \infty }{ \sum _{ m=1 }^{ \infty }{ \frac { 1 }{ { m }^{ 2 }{ (m+n) }^{ 2 } } } }

Ronak Agarwal - 6 years, 3 months ago

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@Ronak Agarwal – I had used Wolfram Alpha to get that summation done , I'm still not well versed with it .

A Former Brilliant Member - 6 years, 3 months ago

Even I have the first pdf , where did you get it from ?

A Former Brilliant Member - 6 years, 3 months ago

I didn't know that you were interested in Naruto ! Do you read the Manga (used to) or watch the Anime ?

A Former Brilliant Member - 6 years, 3 months ago

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Both. I follow it very closely.

Pratik Shastri - 6 years, 3 months ago

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It's good to meet another fan of Naruto !

A Former Brilliant Member - 6 years, 3 months ago

Did the exact same!!

Kartik Sharma - 6 years ago
Ronak Agarwal
Mar 6, 2015

As Pratik Shastri has bought it down to a summation I am here giving a method finding that summation.

I have skipped many small manipulations steps, so kindly go through it a little deeper you will realise what I have done.

Let S = ∑ n = 1 ∞ ∑ m = 1 ∞ 1 ( m + n ) 2 m n S =\displaystyle \sum _{ n=1 }^{ \infty }{ \sum _{ m=1 }^{ \infty }{ \frac { 1 }{ { (m+n) }^{ 2 }mn } } }

Now for m + n = k m+n =k we can rewrite the summation as :

S = ∑ k = 2 ∞ 1 k 2 ∑ m + n = k 1 m n \displaystyle S = \sum _{ k=2 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 } } \sum _{ m+n=k }^{ }{ \frac { 1 }{ mn } } }

Hence S = ∑ k = 2 ∞ 1 k 2 ∑ r = 1 k − 1 1 r ( k − r ) \displaystyle S = \sum _{ k=2 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 } } \sum _{ r=1 }^{ k-1 }{ \frac { 1 }{ r(k-r) } } }

Now S = ( ∑ k = 2 ∞ 1 k 3 ) ∑ r = 1 k − 1 ( 1 r + 1 k − r ) \displaystyle S = (\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ { k }^{ 3 } } )\sum _{ r=1 }^{ k-1 }{ (\frac { 1 }{ r } +\frac { 1 }{ k-r } } ) }

⇒ S = ( ∑ k = 2 ∞ 2 k 3 ) ∑ r = 1 k − 1 ( 1 r ) \Rightarrow \displaystyle S = (\sum _{ k=2 }^{ \infty }{ \frac { 2 }{ { k }^{ 3 } } )\sum _{ r=1 }^{ k-1 }{ (\frac { 1 }{ r } ) } }

⇒ S = 2 ( ∑ k = 2 ∞ 1 k 3 ∑ r = 1 k 1 r − ∑ k = 2 ∞ 1 k 4 ) \Rightarrow \displaystyle S = 2(\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ { k }^{ 3 } } \sum _{ r=1 }^{ k }{ \frac { 1 }{ r } } } -\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ { k }^{ 4 } } } )

⇒ S = 2 ( ∑ k = 1 ∞ 1 k 3 ∑ r = 1 k 1 r − ∑ k = 1 ∞ 1 k 4 ) \Rightarrow \displaystyle S = 2(\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 3 } } \sum _{ r=1 }^{ k }{ \frac { 1 }{ r } } } -\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 4 } } } )

Remember ∑ r = 1 k 1 r = ∑ r = 1 ∞ 1 r − 1 r + k = ∑ r = 1 ∞ k r ( r + k ) \displaystyle \sum _{ r=1 }^{ k }{ \frac { 1 }{ r } } =\sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } -\frac { 1 }{ r+k } } =\sum _{ r=1 }^{ \infty }{ \frac { k }{ r(r+k) } }

Finally our summation becomes :

⇒ S = 2 ( ∑ k = 1 ∞ ∑ r = 1 ∞ 1 k r 2 ( r + k ) − ζ ( 4 ) ) \Rightarrow \displaystyle S =2(\sum _{ k=1 }^{ \infty }{ \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ k{ r }^{ 2 }(r+k) } } } - \zeta(4))

Now since k , r k,r are equivalent and replacing k , r k,r with each other we have :

⇒ S = 2 ( ∑ k = 1 ∞ ∑ r = 1 ∞ 1 r k 2 ( r + k ) − ζ ( 4 ) ) \Rightarrow \displaystyle S = 2(\sum _{ k=1 }^{ \infty }{ \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r{ k }^{ 2 }(r+k) } } } - \zeta(4) )

Add this equation with the previous equation we have :

2 S = 2 ( ∑ k = 1 ∞ ∑ r = 1 ∞ 1 ( r k ) 2 − 2 ζ ( 4 ) ) \displaystyle 2S = 2(\sum _{ k=1 }^{ \infty }{ \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ { (rk) }^{ 2 } } } } - 2\zeta(4))

⇒ S = π 4 180 \Rightarrow S = \dfrac { { \pi }^{ 4 } }{ 180 }

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@Ronak Agarwal ,can you tell me from where i can learn about these technique, I quite a beginnere in multiple integrals. Thanks!

Parth Lohomi - 6 years ago

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I am also a begineer in these, you know making problems is easier than solving them.

Ronak Agarwal - 6 years ago

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OK, I got it thanks

Parth Lohomi - 6 years ago

Hey Ronak! Nice problem! Is it original or did you see it somewhere?

Super Shooter - 5 years, 11 months ago
Akshay Bodhare
Apr 3, 2015

Another method to evaluate the sum, ∑ m = 1 ∞ ∑ n = 1 ∞ 1 m n ( m + n ) 2 = ∑ m = 1 ∞ ∑ n = 1 ∞ m + n m n ( m + n ) 3 \displaystyle\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{mn(m+n)^2}=\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{m+n}{mn(m+n)^3}

∑ m = 1 ∞ ∑ n = 1 ∞ m + n m n ( m + n ) 3 = 2 ∑ m = 1 ∞ ∑ n = 1 ∞ 1 m ( m + n ) 3 \displaystyle\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{m+n}{mn(m+n)^3}=2\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m(m+n)^3}

2 ∑ m = 1 ∞ ∑ n = 1 ∞ 1 m ( m + n ) 3 = 2 ∑ n = 1 ∞ H n ( n + 1 ) 3 \displaystyle2\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac1{m(m+n)^3}=2\sum_{n=1}^{\infty} \frac{H_n}{(n+1)^3}

2 ∑ n = 1 ∞ H n ( n + 1 ) 3 = 2 ( ∑ n = 0 ∞ H n + 1 ( n + 1 ) 3 − ζ ( 4 ) ) \displaystyle2\sum_{n=1}^{\infty} \frac{H_n}{(n+1)^3}=2(\sum_{n=0}^{\infty} \frac{H_{n+1}}{(n+1)^3}-\zeta ({4}))

2 ∑ n = 1 ∞ H n ( n + 1 ) 3 = 2 ( π 4 72 − ζ ( 4 ) ) \displaystyle2\sum_{n=1}^{\infty} \frac{H_n}{(n+1)^3}=2(\frac{\pi^4}{72}-\zeta ({4}))

For the last step, H n = ∑ m = 1 ∞ 1 m − 1 m + n \displaystyle H_n= \sum_{m=1}^{\infty} \frac{1}{m} - \frac{1}{m+n} Therfore, ∑ n = 1 ∞ H n n 3 = ∑ m , n ≥ 1 1 m n 2 ( m + n ) \displaystyle\sum_{n=1}^{\infty} \frac{H_{n}}{n^3}=\sum_{m,n \geq 1} \frac {1}{mn^2(m+n)}

Since the last sum is symmetric,replace m by n and add both,

∑ m , n ≥ 1 1 m n 2 ( m + n ) = 1 2 ζ 2 ( 2 ) \displaystyle\sum_{m,n \geq 1} \frac {1}{mn^2(m+n)}=\frac{1}{2} \zeta^2 (2)

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