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For $x>0$ let $\ f(x) = \int_0^\infty e^{-t-x^2⁄t} t^{-1/2}dt $

the question wants us to show that $\ f(x) = x \int_0^\infty e^{-t-x^2⁄t} t^{-3/2}dt $ by using substitution. However I do not think any substitution works here. What I have done so far is:

  1. found $f'(x)$ as $\ f'(x) = -2x ∫_0^\infty e^{-t-x^2⁄t} t^{-3/2}dt $ and
  2. using integration by parts i have, $\ f(x) = -2x^2 \int_0^\infty e^{-t-x^2⁄t} t^{-3/2}dt $

so that i get $\ f(x)= xf'(x) $ when i solve the differential equation i got $ f(x)= xe^C$ for some positive constant $C$ however the answer should be $ f(x)= Ce^{-2x} $ . I am totally wrong. Please help me. I can not go thorough..

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There are many things that confuse me here. Firstly, neither $xe^C$ nor $Ce^{-2x}$ are solutions to the differential equation $f(x) = x' f(x)$. So I must be misinterpreting you. –  mixedmath Feb 17 '12 at 7:00

1 Answer 1

up vote 3 down vote accepted

You can compute this integral explicitly $$ f(x)=\int\limits_{[0,+\infty)}e^{-t-\frac{x^2}{t}}\frac{dt}{\sqrt{t}}= \left\{s=\frac{x^2}{t}\right\}= \int\limits_{[0,+\infty)}e^{-\frac{x^2}{s}-s}\frac{xds}{\sqrt{s^3}}= \int\limits_{[0,+\infty)}e^{-t-\frac{x^2}{t}}\frac{xdt}{\sqrt{t^3}}= $$ Then $$ f(x)=\frac{1}{2}\left(\int\limits_{[0,+\infty)}e^{-t-\frac{x^2}{t}}\frac{dt}{\sqrt{t}}+\int\limits_{[0,+\infty)}e^{-t-\frac{x^2}{t}}\frac{xdt}{\sqrt{t^3}}\right)= \int\limits_{[0,+\infty)}e^{-t-\frac{x^2}{t}}\left(\frac{1}{2\sqrt{t}}+\frac{x}{2\sqrt{t^3}}\right)dt= $$ $$ \int\limits_{[0,+\infty)}e^{-\left(\sqrt{t}-\frac{x}{\sqrt{t}}\right)^2-2x}d\left(\sqrt{t}-\frac{x}{\sqrt{t}}\right)= \left\{u=\left(\sqrt{t}-\frac{x}{\sqrt{t}}\right)\right\}= \int\limits_{(-\infty,+\infty)}e^{-u^2-2x}du $$ $$ e^{-2x}\int\limits_{(-\infty,+\infty)}e^{-u^2}du=\sqrt{\pi}e^{-2x} $$ So, $$ f'(x)=-2 f(x) $$ P.S. Mathematica gives the same results for $f(x)$

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The equation should be $f'(x) = -2 f(x)$ –  Pedro Tamaroff Feb 17 '12 at 16:31
    
Yes you are right –  Norbert Feb 18 '12 at 9:07

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