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Question Number 133530 by snipers237 last updated on 22/Feb/21

$$I={\int }_0^1\frac{ln(-lnx)}{1+x^2}dx$$

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

$${\int }_0^{\mathrm{\infty }}\frac{log\left(t\right)}{e^{2t}+1}{e}^{t}dt-logx=t\Rightarrow 1=-x\frac{dt}{dx}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{\int }_0^{\mathrm{\infty }}{e}^{-2nt+t}log\left(t\right)dt$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}\frac{1}{2n-1}{\int }_0^{\mathrm{\infty }}{e}^{-u}log\left(u\right)-\frac{log(2n-1)}{2n-1}{\int }_0^{\mathrm{\infty }}{e}^{-u}du$$$$=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n\gamma }{2n+1}-\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{log(2n+1)}{2n+1}$$$$=-\frac{\pi \gamma }{4}+{\mathrm{\Phi }}^{\prime }\left(1\right)$$$$\mathrm{\Phi }\left(\alpha \right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(2n+1)}^{\alpha }}\Rightarrow \mathrm{\Phi }\left(\alpha \right)=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n}log(2n+1)}{{(2n+1)}^{\alpha }}$$

Commented by snipers237 last updated on 23/Feb/21

$$thanksforcryingsir,butithink$$$$ifx\in [0,1]thenicanwrite\frac{1}{1+x^2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-x^2)}^n$$$$butwhent\in [0,\mathrm{\infty }[,wehave{e}^t\in [1,\mathrm{\infty }[,soyou{can}^{\prime }treplacexby{e}^{t}inthepreviousequality$$$$youcouldthingaboutdividingby{e}^{2t}thenumeratorandthedenominatorinthewaytoget$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-t}lnt}{1+e^{-2t}}dtbeforereplacingxby{e}^{-2t}$$$$$$

Commented by Dwaipayan Shikari last updated on 23/Feb/21

$$\frac{e^{-t}log\left(t\right)}{1+e^{-2t}}=\frac{e^{-t}.e^{2t}log\left(t\right)}{e^{2t}+e^{2t-2t}}=\frac{e^{t}log\left(t\right)}{1+e^{2t}}$$

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