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Question Number 217211 by Ghisom last updated on 06/Mar/25

$$\mathrm{a}\mathrm{nice}\mathrm{one}:$$$$\mathrm{prove}\underset{0}{\stackrel{1}{\int }}\sqrt{-\frac{\ln t}{t}}dt=\sqrt{2\pi }$$

Answered by MrGaster last updated on 06/Mar/25

$$\left(1\right):{\int }_0^1\sqrt{-\frac{\ln t}{t}}dt={\int }_0^1{t}^{-\frac{1}{2}}{(-\ln t)}^{\frac{1}{2}}dt$$$$={\int }_0^{\mathrm{\infty }}{e}^{-u}{u}^{-\frac{1}{2}}{\left(\frac{u}{2}\right)}^{\frac{1}{2}}\frac{du}{u}(\mathrm{let}t=e^{-u})$$$$=\frac{1}{\sqrt{2}}{\int }_0^{\mathrm{\infty }}{e}^{-u}{u}^{\frac{1}{2}-1}du$$$$=\frac{1}{\sqrt{2}}\mathrm{\Gamma }\left(\frac{1}{2}\right)$$$$=\frac{1}{\sqrt{2}}\sqrt{\pi }$$$$=\sqrt{\frac{\pi }{2}}$$$$\sqrt{\frac{\pi }{2}}=\frac{\sqrt{\pi }}{\sqrt{2}}=\sqrt{2\pi }$$$$\left(2\right)\mathrm{Let}u=\sqrt{-\ln t}\Rightarrow t=e^{-u^2}\Rightarrow dt=-2{ue}^{-u^2}du$$$${\int }_0^1\sqrt{-\frac{\ln t}{t}}dt={\int }_0^{\mathrm{\infty }}\sqrt{u^2}\cdot (-2{ue}^{-u^2})du=2{\int }_0^{\mathrm{\infty }}{u}^2{e}^{-u^2}du$$$${\int }_0^{\mathrm{\infty }}{e}^{-u^2}du=\frac{\sqrt{\pi }}{2}$$$${\int }_0^{\mathrm{\infty }}{u}^2{e}^{-u^2}du=\frac{1}{2}{\int }_0^{\mathrm{\infty }}u\cdot \left(2{ue}^{-u^2}\right)du=\frac{1}{2}{[-e^{-u^2}]}_0^{\mathrm{\infty }}=\frac{1}{2}$$$$\Rightarrow 2{\int }_0^{\mathrm{\infty }}{u}^2{e}^{-u^2}du2\cdot \frac{1}{2}=1$$$$\therefore {\int }_0^1\sqrt{-\frac{\ln t}{t}}dt=\sqrt{2\pi }$$

Commented by Ghisom last updated on 06/Mar/25

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