Integration Questions

Question Number 173424 by mnjuly1970 last updated on 11/Jul/22

$$\\$$$$\:\:\:\:\mathrm{prove}\:\:\mathrm{that} \\$$$$\\$$$$\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{sinh}\left(\mathrm{x}\right)}\:\mathrm{dx}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{tanh}\left(\frac{\pi}{\mathrm{2}}\right) \\$$$$\\$$

Answered by Mathspace last updated on 11/Jul/22

$$\Psi=\int_{\mathrm{0}} ^{\infty} \:\frac{{sinx}}{{sh}\left({x}\right)}{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\frac{{sinx}}{{e}^{{x}} −{e}^{−{x}} }{dx} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}} {sinx}}{\mathrm{1}−{e}^{−\mathrm{2}{x}} }{dx} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {sinx}\sum_{{n}=\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{nx}} {dx} \\$$$$=\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} {sinx}\:{dx} \\$$$${but}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} {sinxdx} \\$$$$={Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({n}+\mathrm{1}\right){x}+{ix}} {dx}\right) \\$$$${and}\:\int_{\mathrm{0}} ^{\infty} {e}^{\left.−\left({n}+\mathrm{1}\right)+{i}\right){x}} {dx} \\$$$$\left.=\frac{\mathrm{1}}{−\left({n}+\mathrm{1}\right)+{i}}{e}^{\left.−\left({n}+\mathrm{1}\right)+{i}\right){x}} \right]_{\mathrm{0}} ^{\infty} \\$$$$=\frac{\mathrm{1}}{{n}+\mathrm{1}−{i}}=\frac{{n}+\mathrm{1}+{i}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow \\$$$${Im}\left(\int_{\mathrm{0}} ^{\infty} ....\right)=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow \\$$$$\Psi=\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\$$$$=\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}} \\$$$$\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+{i}\right)\left({n}−{i}\right)} \\$$$$=\frac{\Psi\left({i}\right)−\Psi\left(−{i}\right)}{\mathrm{2}{i}}\:{after}\:{we}\:{use} \\$$$$\Psi\left({z}\right)−\Psi\left(\mathrm{1}−{z}\right)=\pi{cotan}\left(\pi{z}\right) \\$$$${or}\:{we}\:{can}\:{developp}\:{cos}\left(\alpha{x}\right) \\$$$${at}\:{fourier}\:{serie}\:{to}\:{find}\:{the}\:{value}... \\$$

Commented by mnjuly1970 last updated on 11/Jul/22

$$\mathrm{grateful}\:\mathrm{sir}\: \\$$

Commented by Tawa11 last updated on 13/Jul/22

$$\mathrm{Great}\:\mathrm{sir} \\$$