Is there an indefinite integral for this?

Calculus Level 4

$\large \displaystyle\int_0^{\infty}\frac{x^4e^x}{(e^x-1)^2} \, dx$

The integral above equals to $\frac ab \pi^4$ for coprime positive integers $a,b$ and that you're given $\displaystyle \sum_{j=1}^\infty \frac1{j^4} = \frac{\pi^4}{90}$ .

Find $a+b$

The answer is 19.

1 solution

Siddharth Bhatt
Jun 12, 2015

Note that the integral can be written as

$\displaystyle\int_0^{\infty} \dfrac{x^4e^{-x}}{(1-e^{-x})^2} dx = \displaystyle\sum_{k=1}^{\infty} k\displaystyle\int_0^{\infty}x^4 e^{-kx} dx = \displaystyle\sum_{k=1}^{\infty}k \cdot \dfrac{24}{k^5} = 24 \zeta(4) = \dfrac4{15} \pi^4$

So $a+b=\boxed{19}$

I did not understand the Riemman Zeta part.. how?

Jun Arro Estrella - 5 years, 5 months ago

elaborate please

Shashank Rustagi - 5 years, 11 months ago

How have you solved. 😒

siddharth bhatt - 5 years, 11 months ago

Basically a binomial expansion which meets with certain coincidence to restore into simple form, I think. A correct direction is crucial.

Lu Chee Ket - 5 years, 6 months ago

Is there an indefinite integral for this?

The answer is 19.

1 solution

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