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I need to find integral of $\int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$. From this Riemann_zeta_function , it gives $\zeta\left( s\right) =\frac{1}{\Gamma\left( s\right) }\int_{0}^{\infty}\frac{x^{s-1}}{e^{x}-1}dx$. Hence for $s=3$ this gives

\begin{align} \int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx & =\zeta\left( 3\right) \Gamma\left( 3\right) \nonumber\\ & =2\zeta\left( 3\right) \tag{1}% \end{align}

My question is, how to use the above result to find integral of $\int% _{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx$? Mathematica gives

\begin{equation} \int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}dx=\frac{3}{2}\zeta\left( 3\right) \tag{2} \end{equation}

I tried to see if there is change of variable or other trick, but I do not see one. I do know about a direct method given in this answer here, but I want to see if it is possible, with some trick, to use the result from (1) to find (2).

Nasser
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  • I am guessing a lot of work has probably been done on that integral with $+$ in the denominator if it is that closely related to $\zeta(3)$. – Countable Jan 28 '21 at 01:08
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    Not necessarily reusing (1) - but if you write $\frac{x^2}{e^x+1} = x^2 e^{-x} - x^2 e^{-2x} + x^2 e^{-3x} - x^2 e^{-4x} + \cdots$, then the integral is $2 - \frac{2}{2^3} + \frac{2}{3^3} - \frac{2}{4^3} = 2(1 + \frac{1}{2^3} + \frac{1}{3^3} + \cdots) - 4(\frac{1}{2^3} + \frac{1}{4^3} + \cdots) = 2 \zeta(3) - \frac{4}{8} \zeta(3)$. – Daniel Schepler Jan 28 '21 at 01:16
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    Oh wait, I suppose if you follow the argument, you could see it becoming: the difference of the two integrals is $2 \int_0^\infty \frac{x^2}{e^{2x}-1}dx$ and then do a change of variables $y = 2x$. – Daniel Schepler Jan 28 '21 at 01:20
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    To generalize to $x^{s-1}$ instead of just $x^2$, the key result is that $\sum_{n=1}^\infty (-1)^n/n^s=\left(1-2^{1-s}\right)\sum_{n=1}^\infty 1/n^s$. – Dispersion Jan 28 '21 at 03:26

2 Answers2

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Let $\displaystyle \mathcal I =\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}\,\mathrm{d}x $ and $\displaystyle \mathcal J = \int_{0}^{\infty}\frac{x^{2}}{e^{x}+1}\,\mathrm{d}x. $

You already know $\mathcal I =2\zeta\left( 3\right)$ but you seek $\mathcal J$.

$\displaystyle \mathcal I =\int_{0}^{\infty}\frac{(2x)^{2}}{e^{2x}-1}\,\mathrm{d}(2x) = \int_0^\infty \frac{8x^2}{e^{2x}-1}\, \mathrm{d}x$

But $\displaystyle \frac{8x^2}{e^{2x}-1} = \frac{4x^2}{e^x-1}-\frac{4x^2}{e^x+1}$

Thus $\displaystyle \mathcal I = 4\int_{0}^{\infty} \frac{x^2}{e^x-1}\,\mathrm{d}x-4\int_{0}^{\infty} \frac{x^2}{e^x+1}\,\mathrm{d}x $

So $\mathcal I = 4 \mathcal I-4 \mathcal J $ and it follows that $\displaystyle \mathcal J =\frac{3}{2}\zeta(3). $

NoName
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Well using the fact that: $$(a+1)(a-1)=a^2-1$$ we can say that: $$\frac{1}{e^{2x}-1}=\frac12\frac{1}{e^x-1}-\frac12\frac{1}{e^x+1}$$ $$\therefore \frac{1}{e^x+1}=\frac{1}{e^x-1}-\frac{2}{e^{2x}-1}$$ so In term of your integral we can say: $$I=\int_0^\infty\frac{x^2}{e^x+1}dx=\int_0^\infty\frac{x^2}{e^{x}-1}dx-2\int_0^\infty\frac{x^2}{e^{2x}-1}dx\tag{1}$$ $$I_1=\int_0^\infty\frac{x^2}{e^x-1}dx\tag{2}$$ Now, the first integral you know the value of so lets look at the second, using the substitution $u=2x\Rightarrow dx=\frac{du}2$ we can say the following: $$\int_0^\infty\frac{x^2}{e^{2x}-1}dx=\int_0^\infty\frac{(u/2)^2}{e^u-1}\frac{du}2=\frac18\int_0^\infty\frac{u^2}{e^u-1}du=\frac18I_1\tag{3}$$

Now from this we can say: $$I=I_1-\frac28I_1=\frac34I_1\tag{4}$$ which is the same as saying: $$\int_0^\infty\frac{x^2}{e^x+1}dx=\frac34\int_0^\infty\frac{x^2}{e^x-1}dx\tag{5}$$

Henry Lee
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