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Question Number 80397 by M±th+et£s last updated on 02/Feb/20

$$showthat$$$${\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}\frac{1}{\sqrt[y]{x^{\pi }}+1}dxdy=2c$$$$whrrecdenotetha{catalan}^{,}sconstant$$

Commented by mathmax by abdo last updated on 02/Feb/20

$$vhangement{}^y\sqrt{x^{\pi }}=tgivex=t^{\frac{y}{\pi }}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{\left({}^y\sqrt{x^{\pi }}\right)+1}={\int }_0^{\mathrm{\infty }}\frac{y}{\pi }{t}^{\frac{y}{\pi }-1}\times \frac{dt}{t+1}=\frac{y}{\pi }{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{y}{\pi }-1}}{t+1}dt$$$$=\frac{y}{\pi }\times \frac{\pi }{sin(\pi .\frac{y}{\pi })}=\frac{y}{siny}\Rightarrow {\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}\frac{dx}{({}^y\sqrt{x^{\pi }}+1)}dy$$$$={\int }_0^{\frac{\pi }{2}}\frac{y}{siny}dy...wecanfindaapproximatvalueforthisintegral$$$$...becontinued...$$

Commented by mathmax by abdo last updated on 03/Feb/20

$$changementtan\left(\frac{y}{2}\right)=tgive$$$${\int }_0^{\frac{\pi }{2}}\frac{ydy}{siny}={\int }_0^1\frac{2arctan\left(t\right)}{\frac{2t}{1+t^2}}\times \frac{2t}{1+t^2}dt=2{\int }_0^{1}arctan\left(t\right)dt$$$$wehave{arctan}^{{}^{\prime }}\left(t\right)=\frac{1}{1+t^2}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{t}^{2n}\Rightarrow$$$$arctan\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}{t}^{2n}+c(c=0)\Rightarrow {\int }_0^{1}arctan\left(t\right)dt$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}{\int }_0^1{t}^{2n}dt=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{{(2n+1)}^2}=K\left(constanteofcatalan\right)$$$$\Rightarrow {\int }_0^{\frac{\pi }{2}}\frac{y}{siny}dy=2K={\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}\frac{dx}{({}^y\sqrt{x^{\pi }}+1)}dy$$

Commented by mathmax by abdo last updated on 03/Feb/20

$$K\sim 0,9159...$$

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