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Question Number 152947 by mnjuly1970 last updated on 03/Sep/21

$$$$$$prove::$$$$\mathit{\varphi }={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{-\frac{1}{x^2}}}{x^4}dx\stackrel{?}{=}\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)$$$$$$

Answered by mindispower last updated on 03/Sep/21

$$=2{\int }_0^{\mathrm{\infty }}\frac{1}{x}{e}^{-\frac{1}{x^2}}.-\frac{1}{2}d\left(\frac{1}{x^2}\right),\frac{1}{x^2}=t$$$$={\int }_0^{\mathrm{\infty }}\sqrt{t}{e}^{-t}dt={\int }_0^{\mathrm{\infty }}{t}^{\frac{3}{2}-t}{e}^{-t}=\mathrm{\Gamma }\left(\frac{3}{2}\right)=\mathrm{\Gamma }(1+\frac{1}{2})$$$$=\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)$$

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