Integration
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Integration
Calculate, if it exists, the integral
\(\displaystyle{\int_{0}^{\infty}\dfrac{x^3}{e^x-1}\,\mathrm{d}x}\)
\(\displaystyle{\int_{0}^{\infty}\dfrac{x^3}{e^x-1}\,\mathrm{d}x}\)
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- Posts: 33
- Joined: Tue May 10, 2016 3:56 pm
Re: Integration
Let the proposed integral be $I$. We show that $I = \pi^4/15$. Indeed,
\begin{eqnarray*}
I & = & \int_0^\infty\,\frac{x^3e^{-x}}{1 - e^{-x}}\,dx\\
& = & \int_0^\infty\,\sum_{n=0}^\infty\,x^3e^{- (n+1)x}\,dx\\
& = & \sum_{n=0}^\infty\,\int_0^\infty\,x^3e^{- (n+1)x}\,dx\\
& = & \sum_{n=0}^\infty\,\frac{1}{(n+1)^4}\,\int_0^\infty\,t^3e^{-t}\,dt\\
& = & 6\,\zeta(4) = \frac{\pi^4}{15}.
\end{eqnarray*}
Here the exchange order of the integral and summation is justified by the monotone convergence theorem, since the summands are all positive.
\begin{eqnarray*}
I & = & \int_0^\infty\,\frac{x^3e^{-x}}{1 - e^{-x}}\,dx\\
& = & \int_0^\infty\,\sum_{n=0}^\infty\,x^3e^{- (n+1)x}\,dx\\
& = & \sum_{n=0}^\infty\,\int_0^\infty\,x^3e^{- (n+1)x}\,dx\\
& = & \sum_{n=0}^\infty\,\frac{1}{(n+1)^4}\,\int_0^\infty\,t^3e^{-t}\,dt\\
& = & 6\,\zeta(4) = \frac{\pi^4}{15}.
\end{eqnarray*}
Here the exchange order of the integral and summation is justified by the monotone convergence theorem, since the summands are all positive.
Re: Integration
Of course we are dealing with a Mellin Transform here. In general it holds that:
$$\int_{0}^{\infty} \frac{x^{s-1}}{e^x-1} \, {\rm d}x = \Gamma(s) \zeta(s) , \; s \in \mathbb{C} \mid s>1$$
And for future reference
$$\int_{0}^{\infty} \frac{x^{s-1}}{e^x+1} \, {\rm d}x = \left\{\begin{matrix}
\Gamma(s) \eta(s) &, & s \in \mathbb{C} \mid s>1 \\
\ln 2& , & s=1
\end{matrix}\right.$$
$$\int_{0}^{\infty} \frac{x^{s-1}}{e^x-1} \, {\rm d}x = \Gamma(s) \zeta(s) , \; s \in \mathbb{C} \mid s>1$$
And for future reference
$$\int_{0}^{\infty} \frac{x^{s-1}}{e^x+1} \, {\rm d}x = \left\{\begin{matrix}
\Gamma(s) \eta(s) &, & s \in \mathbb{C} \mid s>1 \\
\ln 2& , & s=1
\end{matrix}\right.$$
Hidden Message
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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