Let $s>1$. Evaluate the integral:
$$\mathcal{J}=\int_0^\infty \frac{x^{s-1}}{\sinh x} \, {\rm d}x$$
Solution
We note that:
\begin{align*}
\frac{1}{\sinh x}&=\frac{2}{e^x-e^{-x}}\\
&=\frac{2e^x}{e^{2x}-1} \\
&=\frac{2e^x}{e^{2x} \left ( 1- e^{-2x} \right )}\\
&= \frac{2e^{-x}}{1-e^{-2x}}\\
&=2e^{-x} \sum_{k=0}^{\infty} e^{-2kx} \\
&=2 \sum_{k=0}^{\infty}e^{-(2k+1)x}
\end{align*}
Hence,
\begin{align*}
\int_{0}^{\infty}\frac{x^{s-1}}{\sinh x} \, {\rm d}x &= 2\int_{0}^{\infty} x^{s-1} \sum_{k=0}^{\infty} e^{-(2k+1) x } \, {\rm d}x \\
&=2\sum_{k=0}^{\infty}\int_{0}^{\infty}x^{s-1} e^{-(2k+1)x} \, {\rm d}x \\
&= 2 \Gamma (s) \sum_{k=0}^{\infty} \frac{1}{\left ( 2k+1 \right )^s}\\
&= 2 \Gamma (s) \lambda(s)
\end{align*}
Note: The same technic also reveals that, if $a>0$, then:
$$\int_{0}^{\infty}\frac{x^{s-1}}{\sinh ax} \, {\rm d}x = \frac{2}{a^2} \Gamma (s) \lambda(s)$$
$$\mathcal{J}=\int_0^\infty \frac{x^{s-1}}{\sinh x} \, {\rm d}x$$
Solution
We note that:
\begin{align*}
\frac{1}{\sinh x}&=\frac{2}{e^x-e^{-x}}\\
&=\frac{2e^x}{e^{2x}-1} \\
&=\frac{2e^x}{e^{2x} \left ( 1- e^{-2x} \right )}\\
&= \frac{2e^{-x}}{1-e^{-2x}}\\
&=2e^{-x} \sum_{k=0}^{\infty} e^{-2kx} \\
&=2 \sum_{k=0}^{\infty}e^{-(2k+1)x}
\end{align*}
Hence,
\begin{align*}
\int_{0}^{\infty}\frac{x^{s-1}}{\sinh x} \, {\rm d}x &= 2\int_{0}^{\infty} x^{s-1} \sum_{k=0}^{\infty} e^{-(2k+1) x } \, {\rm d}x \\
&=2\sum_{k=0}^{\infty}\int_{0}^{\infty}x^{s-1} e^{-(2k+1)x} \, {\rm d}x \\
&= 2 \Gamma (s) \sum_{k=0}^{\infty} \frac{1}{\left ( 2k+1 \right )^s}\\
&= 2 \Gamma (s) \lambda(s)
\end{align*}
Note: The same technic also reveals that, if $a>0$, then:
$$\int_{0}^{\infty}\frac{x^{s-1}}{\sinh ax} \, {\rm d}x = \frac{2}{a^2} \Gamma (s) \lambda(s)$$
A similar identity for $\frac{1}{\cosh ax}$ is the following:
ReplyDelete$$\int_{0}^{\infty}\frac{x^{s-1}}{\cosh(ax)}dx=\frac{2\Gamma(s)}{(2a)^{s}} \lambda(s)$$