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Question Number 133518 by Abdoulaye last updated on 22/Feb/21

Answered by mnjuly1970 last updated on 22/Feb/21

$$ans:\frac{{\pi }^3}{16}$$

Answered by Dwaipayan Shikari last updated on 22/Feb/21

$${\int }_{-\mathrm{\infty }}^0\frac{t^2{e}^t}{e^{2t}+1}dtlog\left(x\right)=t\Rightarrow 1=x\frac{dt}{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{u^2{e}^u}{1+e^{2u}}dtt=-u$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{\int }_0^{\mathrm{\infty }}{e}^{-2nu+u}{u}^{2}du$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{{(2n-1)}^3}\mathrm{\Gamma }\left(3\right)=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2n+1)}^3}=\frac{{\pi }^3}{16}$$

Commented by Dwaipayan Shikari last updated on 22/Feb/21

$$\frac{\pi }{2}tan\left(\frac{\pi }{2}x\right)=\frac{1}{(1-x)}-\frac{1}{(1+x)}+\frac{1}{(3-x)}-\frac{1}{(3+x)}+...$$$$Differentiatingtwotimesrespecttox$$$$2.\frac{{\pi }^3}{8}{sec}^2\left(\frac{\pi }{2}x\right)tan\left(\frac{\pi }{2}x\right)=\frac{1}{{(1-x)}^3}-\frac{1}{{(1+x)}^3}+\frac{1}{{(3-x)}^3}-\frac{1}{{(3+x)}^3}+..$$$$x=\frac{1}{2}$$$$\frac{{\pi }^3}{32}=1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+...$$

Commented by Abdoulaye last updated on 22/Feb/21

$$what{it}^{\prime }s\mathrm{\Gamma }?$$

Commented by Dwaipayan Shikari last updated on 22/Feb/21

$$Gammafunction$$$$\mathrm{\Gamma }\left(a\right)={\int }_0^{\mathrm{\infty }}{t}^{a-1}{e}^{-t}dt$$

Commented by Abdoulaye last updated on 22/Feb/21

$$thankyouverymuch$$

Commented by Abdoulaye last updated on 23/Feb/21

$$howdidyoufind\frac{\mathrm{\Pi }}{2}tan\left(\frac{\mathrm{\Pi }}{2}x\right)=\frac{1}{(1-x)}-\frac{1}{(1+x)}+\frac{1}{(3-x)}-\frac{1}{(3+x)}+...?$$

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