What about this ?

Calculus Level 5

We know that ∫ 0 ∞ x 2 e x − 1 d x = 2 ζ ( 3 ) \displaystyle \int_0^\infty \dfrac{x^2}{e^x-1}\; dx = 2\zeta(3) . But if we also knew the value of ∫ 0 ∞ ( x e x − 1 ) 2 d x \displaystyle \int_0^\infty \left(\dfrac{x}{e^x-1}\right)^2\; dx , then what would have been the value of:

∫ 0 ∞ x 2 e x − 1 d x + ∫ 0 ∞ ( x e x − 1 ) 2 d x = ? \int_0^\infty \dfrac{x^2}{e^x-1}\; dx + \int_0^\infty \left(\dfrac{x}{e^x-1}\right)^2\; dx = ?

If the answer can be expressed as π A B \dfrac{\pi^A}{B} for positive integers A A and B B then find A × B A\times B .


The answer is 6.

Aditya Narayan Sharma by Aditya Narayan Sharma , 21, India

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2 solutions

First Last
Jun 27, 2017

I = ∫ 0 ∞ x 2 ( e x − 1 ) 2 d x = ∫ 0 ∞ ∑ n = 0 ∞ ∑ i = 0 ∞ x 2 e − x e − n x e − x e − i x d x = ∫ 0 ∞ ∑ n = 1 ∞ ∑ i = 1 ∞ x 2 e − ( n + i ) x d x \displaystyle \text{I}=\int_0^\infty\frac{x^2}{(e^x-1)^2}dx=\int_0^\infty\sum_{n=0}^\infty\sum_{i=0}^\infty x^2e^{-x}e^{-nx}e^{-x}e^{-ix}dx = \int_0^\infty\sum_{n=1}^\infty\sum_{i=1}^\infty x^2e^{-(n+i)x}dx

Using: ∫ 0 ∞ x b e − a x n d x = Γ ( b + 1 n ) n a b + 1 n By subbing u = a x n with Gamma definition \displaystyle\text{Using: }\int_0^\infty x^be^{-ax^n}dx=\frac{\Gamma(\frac{b+1}{n})}{n\,a^{\frac{b+1}{n}}}\text{ By subbing }u=ax^n\text{ with Gamma definition}

∑ n = 1 ∞ ∑ i = 1 ∞ ∫ 0 ∞ x 2 e − ( n + i ) x d x = 2 ∑ n = 1 ∞ ∑ i = 1 ∞ 1 ( n + i ) 3 = 2 [ ∑ n = 1 ∞ ∑ i = 0 ∞ 1 ( n + i ) 3 ] − 2 ζ ( 3 ) \displaystyle\sum_{n=1}^\infty\sum_{i=1}^\infty\int_0^\infty x^2e^{-(n+i)x}dx = 2\sum_{n=1}^\infty\sum_{i=1}^\infty\frac1{(n+i)^3}=2\Bigg[\sum_{n=1}^\infty\sum_{i=0}^\infty\frac1{(n+i)^3}\Bigg]-2\zeta(3)

Looking term by term: 2 ∑ n = 1 ∞ ∑ i = 0 ∞ 1 ( n + i ) 3 = 2 ∑ n = 1 ∞ n n 3 = 2 ζ ( 2 ) \displaystyle\text{Looking term by term: }2\sum_{n=1}^\infty\sum_{i=0}^\infty\frac1{(n+i)^3} = 2\sum_{n=1}^\infty\frac{n}{n^3}=2\zeta(2)

I + 2 ζ ( 3 ) = 2 ζ ( 2 ) − 2 ζ ( 3 ) + 2 ζ ( 3 ) = π 2 3 \text{I}+2\zeta(3) = 2\zeta(2)-2\zeta(3)+2\zeta(3) = \frac{\pi^2}{3}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

A nice formula ∑ n = 1 ∞ ζ ( a , n ) = ζ ( a − 1 ) a > 2 \displaystyle\sum_{n=1}^\infty\zeta(a,n)=\zeta(a-1)\quad a>2

First Last - 3 years, 11 months ago

nice solution!. an interesting recursion arises for the integral ∫ 0 ∞ x k ( e x − 1 ) n d x \int_0^\infty \dfrac{x^k}{(e^x-1)^n} dx

Aareyan Manzoor - 3 years, 11 months ago
Mark Hennings
Jun 28, 2017

Alternatively, ∫ 0 ∞ x 2 ( e x − 1 ) 2 d x = ∫ 0 ∞ x 2 e − 2 x ( ∑ n = 0 ∞ ( n + 1 ) e − n x ) d x = ∑ n = 0 ∞ ( n + 1 ) ∫ 0 ∞ x 2 e − ( n + 2 ) x d x = 2 ∑ n = 0 ∞ n + 1 ( n + 2 ) 3 = 2 ∑ n = 0 ∞ ( 1 ( n + 2 ) 2 − 1 ( n + 2 ) 3 ) = 2 ∑ n = 2 ∞ ( 1 n 2 − 1 n 3 ) = 2 ( ζ ( 2 ) − ζ ( 3 ) ) = 1 3 π 2 − 2 ζ ( 3 ) \begin{aligned} \int_0^\infty \frac{x^2}{(e^x-1)^2}\,dx & = \int_0^\infty x^2 e^{-2x} \left(\sum_{n=0}^\infty (n+1)e^{-nx}\right)\,dx \; = \; \sum_{n=0}^\infty (n+1)\int_0^\infty x^2 e^{-(n+2)x}\,dx \\ & = \; 2\sum_{n=0}^\infty \frac{n+1}{(n+2)^3} \; = \; 2\sum_{n=0}^\infty \left(\frac{1}{(n+2)^2} - \frac{1}{(n+2)^3}\right) \\ & = \; 2\sum_{n=2}^\infty \left(\frac{1}{n^2} - \frac{1}{n^3}\right) \; = \; 2\big(\zeta(2) - \zeta(3)\big) \\ & = \; \tfrac13\pi^2 - 2\zeta(3) \end{aligned} making the answer 2 × 3 = 6 2 \times 3 = \boxed{6} .

6 Helpful 0 Interesting 0 Brilliant 0 Confused

That's exactly how I did :-) , in fact I didn't solve this one but I was solving ∫ 0 ∞ ln ⁡ m ( 1 + x 2 ) x 2 n + 1 d x \displaystyle \int_0^\infty \dfrac{\ln^m(1+x^2)}{x^{2n+1}}\; dx

Aditya Narayan Sharma - 3 years, 11 months ago

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