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The integral has the form, $$\int_0^\infty x^2e^{(-bx^2)}dx$$

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    Really trivial question: set $x=\frac{1}{\sqrt{b}}z$, then $z=\sqrt{t}$. – 2017-02-02

2 Answers 2

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Let $bx^2=u$ or $x=\sqrt{\frac ub}$.

$$\int_0^\infty x^2e^{-bx^2}\ dx=\frac1{2b^{1.5}}\int_0^\infty u^{0.5}e^{-u}\ du$$

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Hint: Evaluate $I(b)=\displaystyle\int_0^\infty e^{-bx^2}~dx,~$ and then notice that your integral can be written as $-I'(b).$