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Question Number 136197 by Ar Brandon last updated on 19/Mar/21

a . Prove that for any real constant a \int_{0}^{\infty} e^{- \frac{a}{x^{2}}} dx = \infty

b . If a and b are real constants, explain why we cannot split the

integral \int_{0}^{\infty} (e^{- \frac{a}{x^{2}}} - e^{- \frac{b}{x^{2}}}) dx as the difference \int_{0}^{\infty} e^{- \frac{a}{x^{2}}} dx - \int_{0}^{\infty} e^{- \frac{b}{x^{2}}} dx

c . If a ⩾ 0 and b ⩾ 0 constants, then prove that

\int_{0}^{\infty} (e^{- \frac{a}{x^{2}}} - e^{- \frac{b}{x^{2}}}) dx = \sqrt{π b} - \sqrt{π a} .

d . If a > b ⩾ 0 constants, then prove that \int_{0}^{\infty} (e^{- \frac{a}{x^{2}}} - e^{- \frac{b}{x^{2}}}) dx = \infty

Answered by Dwaipayan Shikari last updated on 19/Mar/21

I (a) \int_{0}^{\infty} (e^{- \frac{a}{x^{2}}} - e^{- \frac{b}{x^{2}}}) d x

I^{'} (a) = - \int_{0}^{\infty} \frac{e^{- \frac{a}{x^{2}}}}{x^{2}} d x \frac{1}{x} = u \Rightarrow - \frac{1}{x^{2}} = \frac{d u}{d x}

= - \int_{0}^{\infty} e^{- {a u}^{2}} d u = - \frac{1}{2} \sqrt{\frac{π}{a}}

I (a) = - \sqrt{π a} + C \Rightarrow I (b) = - \sqrt{π b} + C = 0 \Rightarrow C = \sqrt{π b}

I (a) = \sqrt{π} (\sqrt{b} - \sqrt{a})