Question Number 222292 by Nicholas666 last updated on 22/Jun/25
$$ \\ $$$$\:\:\:\int_{−\infty} ^{\infty} \mathrm{sech}\left({z}\right)\:\mathrm{sech}\left({z}−{a}\right)\:{dz} \\ $$$$ \\ $$
Answered by Frix last updated on 22/Jun/25
![=4e^a ∫_(−∞) ^(+∞) (e^(2z) /((e^(2z) +1)(e^(2z) +e^(2a) )))dz =^([t=e^(2z) ]) =2e^a ∫_0 ^(+∞) (dt/((t+1)(t+e^(2a) )))= =(1/(sinh a))[ln ((t+1)/(t+e^(2a) ))]_0 ^(+∞) = =((2a)/(sinh a))](https://www.tinkutara.com/question/Q222293.png)
$$=\mathrm{4e}^{{a}} \underset{−\infty} {\overset{+\infty} {\int}}\:\frac{\mathrm{e}^{\mathrm{2}{z}} }{\left(\mathrm{e}^{\mathrm{2}{z}} +\mathrm{1}\right)\left(\mathrm{e}^{\mathrm{2}{z}} +\mathrm{e}^{\mathrm{2}{a}} \right)}{dz}\:\overset{\left[{t}=\mathrm{e}^{\mathrm{2}{z}} \right]} {=} \\ $$$$=\mathrm{2e}^{{a}} \underset{\mathrm{0}} {\overset{+\infty} {\int}}\frac{{dt}}{\left({t}+\mathrm{1}\right)\left({t}+\mathrm{e}^{\mathrm{2}{a}} \right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{sinh}\:{a}}\left[\mathrm{ln}\:\frac{{t}+\mathrm{1}}{{t}+\mathrm{e}^{\mathrm{2}{a}} }\right]_{\mathrm{0}} ^{+\infty} = \\ $$$$=\frac{\mathrm{2}{a}}{\mathrm{sinh}\:{a}} \\ $$