Infinite sine integral

            This is a short post that I want to compute a beautiful integral that will turn to be useful in a future post. It also appears here

I=\int_{0}^{\infty}\frac{sin(sx)}{e^{2\pi x}-1}dx

I=\int_{0}^{\infty}\frac{sin(sx)e^{-2\pi x}}{1-e^{-2\pi x}}dx

I=\int_{0}^{\infty}{sin(sx)e^{-2\pi x}}{\sum_{n=0}^{\infty}e^{-2\pi xn}}dx

I=\sum_{n=0}^{\infty}\int_{0}^{\infty}{sin(sx)e^{-2\pi x(n+1)}}dx

I=\sum_{n=0}^{\infty}\frac{s}{s^{2}+(n+1)^{2} 2^2\pi^{2}}

I=\sum_{n=1}^{\infty}\frac{s}{s^{2}+n^{2} 2^2\pi^{2}}

I=\frac{s}{2^2\pi^2}\sum_{n=1}^{\infty}\frac{1}{\frac{s^{2}}{2^2\pi^2}+n^{2} }

Recall the formula

\boxed{\sum_{n=1}^{\infty}\frac{1}{s^2+n^{2} }=-\frac{1}{2s^2}+\frac{\pi \coth(\pi s)}{2s}}(1)

\frac{s}{2^2\pi^2}\sum_{n=1}^{\infty}\frac{1}{\frac{s^{2}}{2^2\pi^2}+n^{2} }=\frac{s}{2^2\pi^2} \bigg\{-\frac{2 \pi^2}{s^2}+\frac{\pi^2 \coth(\frac{s}{2})}{s} \bigg \}

\frac{s}{2^2\pi^2}\sum_{n=1}^{\infty}\frac{1}{\frac{s^{2}}{2^2\pi^2}+n^{2} }=-\frac{1}{2s}+\frac{\coth(\frac{s}{2})}{4}

\int_{0}^{\infty}\frac{sin(sx)}{e^{2\pi x}-1}dx=-\frac{1}{2s}+\frac{\coth(\frac{s}{2})}{4}(2)

Observe that

\coth \Big(\frac{s}{2} \Big)=\frac{e^{\frac{ s}{2}}+e^{\frac{- s}{2}}}{e^{\frac{ s}{2}}-e^{\frac{- s}{2}}}

=\frac{e^{\frac{ -s}{2}}}{e^{\frac{- s}{2}}}\frac{e^{\frac{ s}{2}}+e^{\frac{- s}{2}}}{e^{\frac{ s}{2}}-e^{\frac{- s}{2}}}

=\frac{1+e^{- s}}{1-e^{- s}}=\frac{1+1-1+e^{- s}}{1-e^{- s}}

=\frac{2}{1-e^{- s}}-\frac{1-e^{- s}}{1-e^{- s}}

\boxed{\coth \Big(\frac{s}{2} \Big)=\frac{2}{1-e^{- s}}-1}(3)

Substituting (3) in (2)

\int_{0}^{\infty}\frac{sin(sx)}{e^{2\pi x}-1}dx= \frac{1}{2}\bigg\{-\frac{1}{s}+\frac{1}{2} \Big(\frac{2}{1-e^{- s}}-1} \Big) \bigg\}


\int_0^{\infty}\frac{\sin (sx)}{e^{2 \pi x}-1}dx=\frac{1}{2}\left(\frac{1}{1-e^{-s}}-\frac{1}{s}-\frac{1}{2} \right)


\boxed{2\int_0^{\infty}\frac{\sin (sx)}{e^{2 \pi x}-1}dx=\frac{1}{1-e^{-s}}-\frac{1}{s}-\frac{1}{2} }

Alternatively

\coth \Big(\frac{s}{2} \Big)=\frac{e^{\frac{ s}{2}}+e^{\frac{- s}{2}}}{e^{\frac{ s}{2}}-e^{\frac{- s}{2}}}

=\frac{e^{\frac{ s}{2}}}{e^{\frac{ s}{2}}}\frac{e^{\frac{ s}{2}}+e^{\frac{- s}{2}}}{e^{\frac{ s}{2}}-e^{\frac{- s}{2}}}

=\frac{e^{ s}+1}{e^{ s}-1}=\frac{e^{ s}+1+1-1}{e^{ s}-1}

=\frac{2}{e^{ s}-1}-\frac{e^{ s}-1}{e^{ s}-1}

\boxed{\coth \Big(\frac{s}{2} \Big)=\frac{2}{e^{ s}-1}+1}(4)

Substituting (4) in (2)

\int_{0}^{\infty}\frac{sin(sx)}{e^{2\pi x}-1}dx= \frac{1}{2}\bigg\{-\frac{1}{s}+\frac{1}{2} \Big(\frac{2}{e^{ s}-1}+1} \Big) \bigg\}

\int_{0}^{\infty}\frac{sin(sx)}{e^{2\pi x}-1}dx= \frac{1}{2}\bigg(\frac{1}{e^{ s}-1}-\frac{1}{s}+ \frac{1}{2} \bigg)

\boxed{2\int_{0}^{\infty}\frac{sin(sx)}{e^{2\pi x}-1}dx= \frac{1}{e^{ s}-1}-\frac{1}{s}+ \frac{1}{2}}(5)

Ricardo Albahari

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