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Question Number 136197 by Ar Brandon last updated on 19/Mar/21
a. Prove that for any real constant a ∫_0 ^∞ e^(−(a/x^2 )) dx=∞ b. If a and b are real constants, explain why we cannot split the integral ∫_0 ^∞ (e^(−(a/x^2 )) −e^(−(b/x^2 )) )dx as the difference ∫_0 ^∞ e^(−(a/x^2 )) dx−∫_0 ^∞ e^(−(b/x^2 )) dx c. If a≥0 and b≥0 constants, then prove that ∫_0 ^∞ (e^(−(a/x^2 )) −e^(−(b/x^2 )) )dx=(√(πb))−(√(πa)). d. If a>b≥0 constants, then prove that ∫_0 ^∞ (e^(−(a/x^2 )) −e^(−(b/x^2 )) )dx=∞
$$\mathrm{a}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{real}\:\mathrm{constant}\:\mathrm{a}\:\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\frac{\mathrm{a}}{\mathrm{x}^{\mathrm{2}} }} \mathrm{dx}=\infty \\ $$$$\mathrm{b}.\:\mathrm{If}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{real}\:\mathrm{constants},\:\mathrm{explain}\:\mathrm{why}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{split}\:\mathrm{the} \\ $$$$\mathrm{integral}\:\:\int_{\mathrm{0}} ^{\infty} \left(\mathrm{e}^{−\frac{\mathrm{a}}{\mathrm{x}^{\mathrm{2}} }} −\mathrm{e}^{−\frac{\mathrm{b}}{\mathrm{x}^{\mathrm{2}} }} \right)\mathrm{dx}\:\mathrm{as}\:\mathrm{the}\:\mathrm{difference}\:\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\frac{\mathrm{a}}{\mathrm{x}^{\mathrm{2}} }} \mathrm{dx}−\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\frac{\mathrm{b}}{\mathrm{x}^{\mathrm{2}} }} \mathrm{dx} \\ $$$$\mathrm{c}.\:\mathrm{If}\:\mathrm{a}\geqslant\mathrm{0}\:\mathrm{and}\:\mathrm{b}\geqslant\mathrm{0}\:\mathrm{constants},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \left(\mathrm{e}^{−\frac{\mathrm{a}}{\mathrm{x}^{\mathrm{2}} }} −\mathrm{e}^{−\frac{\mathrm{b}}{\mathrm{x}^{\mathrm{2}} }} \right)\mathrm{dx}=\sqrt{\pi\mathrm{b}}−\sqrt{\pi\mathrm{a}}. \\ $$$$\mathrm{d}.\:\mathrm{If}\:\mathrm{a}>\mathrm{b}\geqslant\mathrm{0}\:\mathrm{constants},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} \left(\mathrm{e}^{−\frac{\mathrm{a}}{\mathrm{x}^{\mathrm{2}} }} −\mathrm{e}^{−\frac{\mathrm{b}}{\mathrm{x}^{\mathrm{2}} }} \right)\mathrm{dx}=\infty \\ $$
Answered by Dwaipayan Shikari last updated on 19/Mar/21
I(a)∫_0 ^∞ (e^(−(a/x^2 )) −e^(−(b/x^2 )) )dx I′(a)=−∫_0 ^∞ (e^(−(a/x^2 )) /x^2 )dx (1/x)=u⇒−(1/x^2 )=(du/dx) =−∫_0 ^∞ e^(−au^2 ) du=−(1/2)(√(π/a)) I(a)=−(√(πa))+C⇒I(b)=−(√(πb))+C=0⇒C=(√(πb)) I(a)=(√π)((√b)−(√a))
$${I}\left({a}\right)\int_{\mathrm{0}} ^{\infty} \left({e}^{−\frac{{a}}{{x}^{\mathrm{2}} }} −{e}^{−\frac{{b}}{{x}^{\mathrm{2}} }} \right){dx} \\ $$$${I}'\left({a}\right)=−\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\frac{{a}}{{x}^{\mathrm{2}} }} }{{x}^{\mathrm{2}} }{dx}\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{x}}={u}\Rightarrow−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\frac{{du}}{{dx}} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} {e}^{−{au}^{\mathrm{2}} } {du}=−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{a}}}\: \\ $$$${I}\left({a}\right)=−\sqrt{\pi{a}}+{C}\Rightarrow{I}\left({b}\right)=−\sqrt{\pi{b}}+{C}=\mathrm{0}\Rightarrow{C}=\sqrt{\pi{b}} \\ $$$${I}\left({a}\right)=\sqrt{\pi}\left(\sqrt{{b}}−\sqrt{{a}}\right) \\ $$

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