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Question Number 85781 by Joel578 last updated on 24/Mar/20

$${\int }_0^1{\left(\ln \frac{1}{x}\right)}^{-3/2}dx$$

Commented by Joel578 last updated on 24/Mar/20

$$\mathrm{If}\mathrm{we}\mathrm{use}u=\ln \frac{1}{x},\mathrm{the}\mathrm{integral}\mathrm{will}\mathrm{become}$$$${\int }_0^{\mathrm{\infty }}{u}^{-3/2}{e}^{-u}du\mathrm{which}\mathrm{is}\mathrm{\Gamma }(-\frac{1}{2})=-2\sqrt{\pi }$$$$\mathrm{But}\mathrm{we}\mathrm{know}\mathrm{for}0<x<1,\mathrm{the}\mathrm{graph}\mathrm{of}$$$$y={\left(\ln \frac{1}{x}\right)}^{-3/2}\mathrm{is}\mathrm{always}\mathrm{above}x-\mathrm{axis},\mathrm{so}$$$$\mathrm{the}\mathrm{integral}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{be}\mathrm{negative}.$$$$\mathrm{I}s\mathrm{my}\mathrm{method}\mathrm{invalid}?$$

Commented by MJS last updated on 24/Mar/20

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure},\mathrm{but}\mathrm{the}\mathrm{integral}\mathrm{\Gamma }\left(x\right)\mathrm{only}$$$$\mathrm{converges}\mathrm{for}x>0.\mathrm{for}x<0\wedge x\notin \mathbb{Z}\mathrm{we}\mathrm{must}$$$$\mathrm{calculate}\mathrm{backwards}:\mathrm{\Gamma }(-\frac{1}{2})=\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{-\frac{1}{2}}=-2\sqrt{\pi }$$$$\mathrm{so}\mathrm{the}\mathrm{problem}\mathrm{might}\mathrm{be},\mathrm{if}\mathrm{you}\mathrm{want}\mathrm{to}$$$$\mathrm{integrate}{\left(\ln \frac{1}{x}\right)}^{-\frac{3}{2}}\mathrm{you}\mathrm{cannot}\mathrm{use}\mathrm{the}$$$$\mathrm{gamma}\mathrm{function}$$

Answered by MJS last updated on 24/Mar/20

$$\int {\left(\ln \frac{1}{x}\right)}^{-3/2}dx=$$$$[t=\ln \frac{1}{x}\to dx=-xdt]$$$$=-\int {\mathrm{e}}^{-t}{t}^{-3/2}dt=$$$$[\mathrm{by}\mathrm{parts}:f{}^{\prime }=t^{-3/2}\to f=-2t^{-1/2}$$$$g={\mathrm{e}}^{-t}\to g^{\prime }=-{\mathrm{e}}^{-t}]$$$$={2\mathrm{e}}^{-t}{t}^{-1/2}+2\int {\mathrm{e}}^{-t}{t}^{-1/2}dt=$$$$$$$$[2\int {\mathrm{e}}^{-t}{t}^{-1/2}dt=$$$$[u=\sqrt{t}\to dt=2\sqrt{t}du]$$$$=4\int {\mathrm{e}}^{-u^2}du=2\sqrt{\pi }\int \frac{{2\mathrm{e}}^{-u^2}}{\sqrt{\pi }}du=2\sqrt{\pi }\mathrm{erf}\left(u\right)=$$$$=2\sqrt{\pi }\mathrm{erf}\left(\sqrt{t}\right)]$$$$$$$$={2\mathrm{e}}^{-t}{t}^{-1/2}+2\sqrt{\pi }\mathrm{erf}\left(\sqrt{t}\right)=$$$$=\frac{2x}{\sqrt{\ln \frac{1}{x}}}+2\sqrt{\pi }\mathrm{erf}\left(\sqrt{\ln \frac{1}{x}}\right)+C$$$$\mathrm{but}\mathrm{this}\mathrm{integral}\mathrm{does}\mathrm{not}\mathrm{converge}\mathrm{for}$$$$0⩽x⩽1...$$

Commented by Joel578 last updated on 25/Mar/20

$$thankyouverymuch$$

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