1
$\begingroup$

Could somebody explain the processes in the following integral solution?

$$\int_{0}^{\infty}f(x)dx, $$

$\text{ where } f(x) = \frac{x^{3}}{\theta^{2}}e^{-x^{2}/(2\theta^{2})}$

$$\int_{0}^{\infty}\frac{x^{3}}{\theta^{2}}e^{-x^{2}/(2\theta^{2})}dx \dots (1)$$

$$= \int_{0}^{\infty}x^{2}e^{-x^{2}/(2\theta^{2})}d\frac{x^{2}}{2\theta^{2}} \dots (2)$$

$$= 2\theta^{2}\int_{0}^{\infty}ye^{-y}dy \dots (3)$$

$$= 2\theta^{2}\Gamma(2) \dots (4)$$

$$= 2\theta^{2} \dots (5)$$

More specifically, whilst it seems that the substitution $y = x^{2}/2\theta^{2}$ is made in line 2 (and made further explicit in line 3), why does the $x^{3}$ term become $x^{2}$ and why does the $\theta^{2}$ term in the denominator vanish?

Finally, how does the integral in line 3 evaluate to the gamma function with an argument of 2?

$\endgroup$
2
$\begingroup$

The main idea is to change the variable $x$ in such a way to simplify the integral. In this case it will be succeeded if the exponent has the simplest power. So in step $(1)$ variable $x^3$ is split into $x^2$ (which you see in step $(2)$) and $x$ which is carried under the differential sign to get $x^2$ under it, because $\,{\rm d}{x^2}=2x\,{\rm d}x$. $\theta$ is a constant, so it goes under the differential sign too. Then the change of variable is performed: $y=\frac{x^2}{2\theta}$ (for the new variable the integration bounds do not change). In step $(3)$ we see the definition of the Gamma function: $ \Gamma(z) = \int_0^\infty y^{z-1} e^{-y}\,{\rm d}y$ with $z=2$. So the answer is $\Gamma(2)$.

$\endgroup$
  • $\begingroup$ Right, but a messy way of doing it. Just say you want to do $y = x^2 / 2 \theta^2$, carry out the variable change. And why the useless detour though $\Gamma$, the last integral is easy enough to integrate by parts. I'd go looking for a clearer text... $\endgroup$ – vonbrand Mar 31 '13 at 15:58
  • $\begingroup$ I had misunderstood the substitution principle, but that's cleared up any confusion. Thanks for that! $\endgroup$ – Will Clyne Mar 31 '13 at 16:05
  • $\begingroup$ @vonbrand Presented formulae are quite explicit, so if Will Clyne did not understand them, then the "messy" job should be done to help understand EVERY small step. $\endgroup$ – Caran-d'Ache Mar 31 '13 at 16:06
0
$\begingroup$

Doesn't answer the question, but I'd do: $$ \int_0^\infty x^3 e^{- x^2 / 2 \theta^2} dx $$ As Caran-d'Ache says, try to simplify the exponent. Try $y = x^2 / 2 \theta^2$, that is $x = \theta \sqrt{2 y}$. When $x = 0$, $y = 0$; $x = \infty$ gives $y = \infty$. So : $$ \begin{align*} d x &= \theta \sqrt{2} \frac{d y}{\sqrt{y}} \\ x^3 e^{- x^2 / 2 \theta^2} &= \left( \theta \sqrt{2 y} \right)^3 e^{-y} \\ \int_0^\infty x^3 e^{- x^2 / 2 \theta^2} dx &= \int_0^\infty \left( \theta \sqrt{2 y} \right)^3 e^{-y} \cdot \theta \sqrt{2} \frac{d y}{\sqrt{y}} \\ &= 2 \theta \int_0^\infty y e^{-y} d y \end{align*} $$ For the remaining integral, use integration by parts, $u = y$, $d v = e^{-y} dy$, so $du = dy$, $v = - e^{-y}$: $$ \int_0^\infty y e^{-y} dy = \left. - y e^{-y} \right|_0^\infty + \int_0^\infty e^{-y} d y = 1 $$ And the final result is: $$ \int_0^\infty x^3 e^{- x^2 / 2 \theta^2} dx = 2 \theta $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.