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Question Number 139524 by mnjuly1970 last updated on 28/Apr/21

$$......advancedcalculus.....$$$$provethat:$$$$i::\mathit{\varphi }={\int }_0^{\mathrm{\infty }}(\frac{1}{2}{e}^{-2x}-\frac{1}{1+e^x})\frac{1}{x}dx=log\left(\frac{1}{\sqrt{\pi }}\right)$$$$ii::{\int }_0^{\mathrm{\infty }}(\frac{1}{2}-\frac{1}{1+e^{-x}})\frac{e^{-2x}}{x}dx=log\left(\frac{\sqrt{\pi }}{2}\right)$$

Answered by mindispower last updated on 30/Apr/21

$$i..iiaresimilarinwaythiscansolveby$$$$let\mathrm{\varnothing }={\int }_0^{\mathrm{\infty }}(\frac{e^{-2x}}{2}-\frac{e^{-x}}{1+e^{-x}}).\frac{dx}{x}$$$$e^{-x}=t\Rightarrow \varphi ={\int }_0^1(\frac{1}{1+t}-\frac{t}{2})\frac{dt}{ln\left(t\right)}$$$$letf\left(z\right)={\int }_0^1(\frac{1}{1+t}-\frac{t}{2})\frac{t^z}{ln\left(t\right)}dt$$$$f^{\prime }\left(z\right)={\int }_0^1(\frac{t^z}{1+t}-\frac{t^{1+z}}{2})dt=-\frac{1}{2(2+z)}+{\int }_0^1\frac{t^z-t^{1+z}}{1-t^2}dt$$$$t^2\to tin2nd\iff f^{\prime }\left(z\right)=-\frac{1}{2(2+z)}+\frac{1}{2}(-\gamma +{\int }_0^1\frac{1-t^{\frac{z}{2}}}{1-t}.dt)$$$$-\frac{1}{2}(-\gamma +{\int }_0^1\frac{1-t^{\frac{z-1}{2}}}{1-t}dt)$$$$\mathrm{\Psi }(x+1)=-\gamma +{\int }_0^1\frac{1-t^x}{1-t}dt\Rightarrow$$$$f^{\prime }\left(z\right)=-\frac{1}{2(2+z)}+\frac{1}{2}(\mathrm{\Psi }(\frac{z}{2}+1)-\mathrm{\Psi }\left(\frac{z+1}{2}\right))$$$$f\left(z\right)=-\frac{1}{2}ln(2+z)+log\left(\frac{\mathrm{\Gamma }(\frac{z}{2}+1)}{\mathrm{\Gamma }\left(\frac{z+1}{2}\right)}\right)+c$$$$\mathrm{\Gamma }(z+a)\sim \sqrt{2\pi }.z^{z+a-\frac{1}{2}}{e}^{-z}$$$$\frac{\mathrm{\Gamma }(\frac{z}{2}+1)}{\mathrm{\Gamma }\left(\frac{z+1}{2}\right)}\sim \frac{{\left(\frac{z}{2}\right)}^{\frac{z}{2}+\frac{1}{2}}}{{\left(z\right)}^{\frac{z}{2}}}$$$$f\left(z\right)\sim ln(\frac{1}{\sqrt{2+z}}.\sqrt{\frac{z}{2}})+c\Rightarrow \underset{z\to \mathrm{\infty }}{\lim }f\left(z\right)=ln\left(\frac{1}{\sqrt{2}}\right)+c$$$$backtof\left(z\right)={\int }_0^1(\frac{1}{1+t}-\frac{t}{2})\frac{t^z}{ln\left(t\right)}dt\Rightarrow \underset{z\to \mathrm{\infty }}{\lim }f\left(z\right)=0$$$$\Rightarrow c=ln\left(\sqrt{2}\right)\iff f\left(z\right)={\int }_0^1(\frac{1}{1+t}-\frac{t}{2}).\frac{t^z}{ln\left(t\right)}=ln\left(\frac{1}{\sqrt{2+z}}\right)$$$$+log\left(\frac{\mathrm{\Gamma }(\frac{z}{2}+1)}{\mathrm{\Gamma }\left(\frac{z+1}{2}\right)}\right)+ln\left(\sqrt{2}\right)$$$$f\left(0\right)=\varphi =ln\left(\frac{1}{\sqrt{2}}\right)+ln\left(\sqrt{2}\right)+log\left(\frac{\mathrm{\Gamma }\left(1\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}\right)=log\left(\frac{1}{\sqrt{\pi }}\right)$$$$\iff \varphi ={\int }_0^{\mathrm{\infty }}(\frac{e^{-2x}}{2}-\frac{e^{-x}}{1+e^{-x}}).\frac{dx}{x}=ln\left(\frac{1}{\sqrt{\pi }}\right)$$$$$$$$$$$$$$

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