Question

نشان دهید که: $$I= \int _ {- \infty } ^ { \infty } 1- e^{- \frac{a}{ x^{2} } } dx=2 \sqrt{ \pi a} $$

Comments

$$I(a)= \int _ {- \infty } ^ { \infty } 1- e^{ \frac{-a}{ x^{2} } } dx=2 \int _0^ \infty 1- e^{ \frac{-a}{ x^{2} } } dx \wedge u= \frac{1}{x}  \Longrightarrow I(a)=2 \int _ \infty ^0 (1- e^{-a u^{2} } ) \frac{-du}{ u^{2} }  \Longrightarrow I'(a)=2 \int _0^ \infty  e^{-a u^{2} } du \wedge  t^{2} =a u^{2}  \Longrightarrow I'(a)=2 \int _0^ \infty  e^{- t^{2} }  \frac{dt}{ \sqrt{a} } = \frac{a}{b}  \int _ {- \infty } ^ { \infty }  e^{ -t^{2} } dt= \sqrt{ \frac{ \pi }{a} } = \sqrt{ \pi }  a^{ \frac{-1}{2} }  \Longrightarrow I(a)=2 \sqrt{ \pi a} +k \wedge k=0 \Longrightarrow I(a)=2 \sqrt{ \pi a}$$

Answer 1

$$I(a)= \int _ {- \infty } ^ { \infty } 1- e^{ \frac{-a}{ x^{2} } } dx=2 \int _0^ \infty 1- e^{ \frac{-a}{ x^{2} } } dx \wedge u= \frac{1}{x} \Longrightarrow x= \frac{1}{u} \Longrightarrow dx= -\frac{du}{ u^{2} } \Longrightarrow I(a)=2 \int _ \infty ^0 (1- e^{-a u^{2} } ) \frac{-du}{ u^{2} } \Longrightarrow I(a)=2 \int _0^ \infty \frac{1- e^{-a u^{2} } }{ u^{2} } \Longrightarrow I'(a)=2 \int _0^ \infty e^{-a u^{2} } du \wedge t^{2} =a u^{2} \Longrightarrow t=u \sqrt{a} \Longrightarrow dt= \sqrt{a} du \Longrightarrow I'(a)=2 \int _0^ \infty e^{- t^{2} } \frac{dt}{ \sqrt{a} } = \frac{a}{b} \int _ {- \infty } ^ { \infty } e^{ -t^{2} } dt= \sqrt{ \frac{ \pi }{a} } = \sqrt{ \pi } a^{ \frac{-1}{2} } \Longrightarrow I(a)=2 \sqrt{ \pi a} +k \wedge k=0 \wedge a=0 \wedge I(0)=0 \Longrightarrow I(a)=2 \sqrt{ \pi a}$$