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How do I integrate this ?

$$\int^{ \infty }_0\frac{x^4}{e^{x^2}+1}dx$$

I just brute force integrated it , but didn't get a valid answer . I'm sure it must be done by some properties of definite integral , I'm not sure what properties though.

My attempt : https://ibb.co/b422TH

In that attempt , $1+e^{-t}=m\implies -e^{-t}dt=dm$ Minus sign was missing , but even after I corrected it and then integrated my final expression by parts , I'm still getting answer as not defined $(\infty)$

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    $\begingroup$ What does "I just brute force integrated it , but didn't get a valid answer" really mean? I do not see how a numerical integration may lead to an incorrect answer. Anyway, such integral is clearly related with $\zeta\left(\frac{5}{2}\right)$. Enforce the substitution $x=\sqrt{z}$, expand $\frac{1}{e^z+1}$ as a geometric series, exploit $\int_{0}^{+\infty}x^{\alpha}e^{-x}\,dx = \Gamma(\alpha+1)$. $\endgroup$ – Jack D'Aurizio Feb 24 '18 at 14:53
  • $\begingroup$ @JackD'Aurizio Not sure what you mean $\endgroup$ – Tanuj Feb 24 '18 at 14:54
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    $\begingroup$ Please show your actual attempts and provide more context. $\endgroup$ – Jack D'Aurizio Feb 24 '18 at 14:55
  • $\begingroup$ By brute force I meant , i did not use any definite integral properties first , I simplified it and then put the limits. $\endgroup$ – Tanuj Feb 24 '18 at 14:58
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    $\begingroup$ @KingTut: $$\zeta(s)=\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x-1}\,dx$$ $\endgroup$ – Jack D'Aurizio Feb 24 '18 at 15:21
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$$A=\int_{0}^{\infty}\frac{x^4}{e^{x^2}+1} dx = \int_{0}^{\infty} \frac{x^4 e^{-x^2}}{1+e^{-x^2}} dx\\ =\int_{0}^{\infty}\sum_{n=0}^{\infty}{x^4 (-1)^ne^{-(n+1)x^2}} dx$$

In the interval, $[0, \infty]$, For all $n \in \mathbb{N}$, $|(-1)^n x^4 e^{- (n+1) x^2}| \le x^4 e^{- (n+1) x^2}$ and $\sum_{n=0}^{\infty}{x^4 e^{- (n+1) x^2}}=\frac{x^4}{e^{x^2}-1}$ which is bounded function and$\sum_{n=0}^{\infty}\int_{0}^{\infty}{x^4 e^{- (n+1) x^2}}dx = \frac{3 \sqrt{\pi}}{8}\zeta(\frac{5}{2}) < + \infty$.

Therefore, we could interchange integration and summation.

$$A=\sum_{n=0}^{\infty} (-1)^n \int_{0}^{\infty}{x^4 e^{-(n+1)x^2}} dx\\ = \sum_{n=0}^{\infty} (-1)^n\int_{0}^{\infty}{\frac{1}{2} t^{3/2} e^{-(n+1)t}} dt\\ = \sum_{n=0}^{\infty} (-1)^n \frac{3 \sqrt{\pi}}{8}\frac{1}{(1+n)^{5/2}}\\ = \frac{3 \sqrt{\pi}}{8}\sum_{n=0}^{\infty} (-1)^n \frac{1}{(1+n)^{5/2}}\\ = \frac{3 \sqrt{\pi}}{8}\eta(\frac{5}{2})\\ = \frac{3 \sqrt{\pi}}{8}(1-2^{-3/2})\zeta(\frac{5}{2}) $$

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  • $\begingroup$ I didn't get that second step . How do you convert that into sigma ? $\endgroup$ – Tanuj Feb 24 '18 at 15:19
  • $\begingroup$ Oh.. I make some mistakes in first step. $\endgroup$ – ChoMedit Feb 24 '18 at 15:21
  • $\begingroup$ @Tanuj I just use series expansion. $\endgroup$ – ChoMedit Feb 24 '18 at 15:23
  • $\begingroup$ For each x in the interval(except 0), it converges to original function because it is geometric series. $\endgroup$ – ChoMedit Feb 24 '18 at 15:35

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