This question is from an old qualification exam for a master degree in my college. I tried to solve it as prep.

For $x>0$m let $f(x) = \int_{0}^\infty e^{-t-\frac{x^2}{t}}\cdot t^{-\frac{1}{2}}dt $.


Using a substitution, show that $$f\left(x\right)= x\int_{0}^\infty e^{-t-\frac{x^2}{t}}\cdot t^{-\frac{3}{2}} dt$$


Then show that $f(x) = C \cdot e^{-2x}$ for some positive constant $C$.

For part 1, I substitute $\frac{x^2}{t}$ as u, and proved $f\left(x\right)= x\int_{0}^\infty e^{-u-\frac{x^2}{u}}\cdot u^{-\frac{3}{2}} du$

Please help me to solve the second part of the question.


2 Answers 2


For part $1)$, as has already been said, just substitute $u = \frac{x^2}{t}$

For part $2)$, differentiate $f(x)$ w.r.t. $x$ in the original definition of $f(x)$, and since the limits are constants not dependent on $x$, you get:

$f'(x) = \int_0^{\infty} e^{-t-\frac{x^2}{t} } \cdot \left( -\frac{2x}{t}\right) \cdot t^{-\frac{1}{2}}dx = (-2)\left(x\int_0^{\infty} e^{-t-\frac{x^2}{t} } \cdot t^{-\frac{3}{2}}dx\right) = (-2) f(x)$ which gives you a differential equation in f(x), that is:

$f'(x) = -2f(x) $,

whose solution is as given.

  • $\begingroup$ I guess what you finally get is $-2f(x)$, not $-2xf(x)$. $\endgroup$ Mar 13, 2016 at 9:01
  • 1
    $\begingroup$ @SangchulLee My bad! Thanks for pointing it out! I forgot about the extra factor of $x$ that appears in the right hand side of the equation of the first part. It has been corrected. $\endgroup$ Mar 13, 2016 at 11:39

For part 1), we can use the substitution $u = x^2/t$ (and accordingly $dt = -(x/u)^2 \, du$) to write

$$ f(x) = \int_{0}^{\infty} e^{-\frac{x^2}{u} - u} \left(\frac{x^2}{u}\right)^{-1/2} \cdot \frac{x^2}{u^2} \, du = x \int_{0}^{\infty} e^{-u-\frac{x^2}{u}} u^{-3/2} \, du. $$

For part 2), we use part 1) to rewrite $f(x)$ as

$$ f(x) = \frac{f(x) + f(x)}{2} = \int_{0}^{\infty} e^{-t - \frac{x^2}{t}} \left( \frac{ t^{-1/2} + xt^{-3/2}}{2} \right) \, dt. $$

Here the crucial ideal is to introduce the substitution $v = t^{1/2} - xt^{-1/2}$. As $t$ moves from 0 to $\infty$, this new variables $v =v(t)$ sweeps all the possible values in $\Bbb{R}$ from $-\infty$ to $\infty$. Also we easily notice that $v^2 + 2x = t + (x^2/t)$ and $dv = \frac{1}{2}(t^{-1/2} + xt^{-3/2})$. Thus it follows that

$$ f(x) = \int_{-\infty}^{\infty} e^{-(v^2 + 2x)} \, dv = \sqrt{\pi}e^{-2x}. $$

Remark. Aside from the computation itself, here is a possible source of motivation of this problem. If you have some knowledge on Brownian motion, you can check that this function is closely related to the hitting time of the Brownian motion as follows: If $B = (B_t : t \geq 0)$ is the standard 1-dim Brownian motion started at 0 and $T_a = \inf\{ t \geq 0 : B_t = a \}$ is the hitting time at level $a > 0$, then

$$ \Bbb{P}(T_a \in dt) = \frac{a}{\sqrt{2\pi}} e^{-a^2/2t} t^{-3/2} \, dt. \tag{1} $$

Consequently the Laplace transform of $T_a$ is

$$ \Bbb{E}e^{-sT_a} = \int_{0}^{\infty} e^{ - st} \, \Bbb{P}(T_a \in dt) = e^{-\sqrt{2s}a}. \tag{2} $$

These two facts can be obtained by using the reflection principle and the exponential martingale trick, respectively. That said, in principle we can prove 2) without any direct computation. (This is certainly a sledgehammer method, though.)

  • $\begingroup$ For the part 1, I have a question on calculation. We can get $dt = -\left(x/u\right)^2 du$ and put it into the integrand. How did you cancel the minus sign of it? $\endgroup$
    – Nhay
    Mar 13, 2016 at 21:29
  • 1
    $\begingroup$ @NoboruHayashi, That sign is cancelled out by inverting the lower bound and the upper bound. Under the substitution, the order is changed from $0\to\infty$ to &\infty\to0$, and the negative sign is used to invert it. $\endgroup$ Mar 13, 2016 at 21:37
  • $\begingroup$ Now I fully understand this problem. Thank you very much! $\endgroup$
    – Nhay
    Mar 13, 2016 at 21:39
  • $\begingroup$ @NoboruHayashi You're welcome :) $\endgroup$ Mar 13, 2016 at 21:40

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