Question Number 151144 by mnjuly1970 last updated on 18/Aug/21
Answered by qaz last updated on 18/Aug/21
$$\frac{\mathrm{1}}{\mathrm{n}}\int_{−\infty} ^{+\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{x}} \right)}{\mathrm{cosh}\:\left(\frac{\mathrm{x}}{\mathrm{n}}\right)}\mathrm{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{nx}} \right)}{\mathrm{cosh}\:\mathrm{x}}\mathrm{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{e}^{−\mathrm{nx}} \right)}{\mathrm{cosh}\:\mathrm{x}}\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\int_{−\infty} ^{+\infty} \frac{\mathrm{dx}}{\mathrm{cosh}\:\mathrm{x}} \\ $$$$=\pi\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{2x}} }\mathrm{dx} \\ $$$$=−\pi\mathrm{tan}^{−\mathrm{1}} \mathrm{e}^{−\mathrm{x}} \mid_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$
Commented by mnjuly1970 last updated on 18/Aug/21
$${excellent}\:{master}\:{bravo}… \\ $$