Question Number 211528 by mnjuly1970 last updated on 12/Sep/24
$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{tanh}^{−\mathrm{1}} \:\left({x}^{\mathrm{2}} \:\right)}{{x}^{\:\mathrm{2}} }\:{dx}=\:?\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Frix last updated on 12/Sep/24
$$\mathrm{Using}\:\mathrm{Feynman}'\mathrm{s}\:\mathrm{Technique} \\ $$$${f}\left(\alpha\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{tanh}^{−\mathrm{1}} \:\left(\alpha{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\mathrm{I}\:\mathrm{get}\:{I}={f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 12/Sep/24
$$\: \\ $$