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Question Number 131875 by mnjuly1970 last updated on 09/Feb/21

$$...nicecalculus..$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{ln\left(cosh\left(x\right)\right)}{cosh\left(x\right)}dx=???$$

Answered by mindispower last updated on 09/Feb/21

$$={\int }_1^{\mathrm{\infty }}\frac{ln\left(t\right)dt}{t\sqrt{t^2-1}}$$$$={\int }_1^{\mathrm{\infty }}\frac{ln\left(t\right).tdt}{t^2\sqrt{t^2-1}}=\frac{1}{4}{\int }_1^{\mathrm{\infty }}\frac{ln\left(t\right)dt}{t\sqrt{t-1}}$$$$t=\frac{1}{y}\Rightarrow \frac{1}{4}{\int }_0^1\frac{-ln\left(y\right)dy}{\sqrt{y}.\sqrt{1-y}}$$$$=-\frac{1}{4}{\int }_0^{1}ln\left(y\right)y^{\frac{1}{2}-1}{(1-y)}^{\frac{1}{2}-1}dy$$$$\beta (a,b)={\int }_0^1{t}^{a-1}{(1-t)}^{b-1}dt$$$$wefind=-\frac{1}{4}{\partial }_a\beta (\frac{1}{2},\frac{1}{2})=-\frac{1}{4}\beta (\frac{1}{2},\frac{1}{2})[\mathrm{\Psi }\left(\frac{1}{2}\right)-\mathrm{\Psi }\left(1\right)]$$$$=-\frac{\pi }{4}[-2ln\left(2\right)]=\frac{\pi ln\left(2\right)}{2}$$

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