Dealing with undefined numbers in definite integral

Question

While solving this particular definite integral: $$\int_0^\infty x^2 e^{-2ax} dx$$

I could evaluate the indefinite integral as: $$- \frac{e^{-2ax} (2a^2x^2 + 2ax + 1)}{4a^3}$$

But I fall into this problem of $0 * \infty$ upon putting the limits. What's the correct way to evaluate this integral?

Answer 1

As mentioned in the comments and answers, it is clear that you must have $a > 0$ for the integral to converge. Now, if you want to evaluate the integral as you suggest, separate the terms \begin{align*} - \frac{e^{-2ax} (2a^2x^2 + 2ax + 1)}{4a^3} \Big|_{0}^{\infty} & = \left(\lim_{x \to \infty} - \frac{e^{-2ax} (2a^2x^2 + 2ax + 1)}{4 a^3} \right) + \frac{1}{4a^{3}} \\ & = -\left(\lim_{x \to \infty} \frac{e^{-2ax}2a^2x^2}{4 a^3} + \frac{2e^{-2ax}ax}{{4 a^3} } + \frac{e^{-2ax}}{4 a^3} \right) + \frac{1}{4a^{3}} \\ & = -\left(\lim_{x \to \infty} \frac{x^2}{2 e^{2ax}a} + \frac{x}{{2e^{2ax} a^2} } + \frac{1}{4e^{2ax} a^3} \right) + \frac{1}{4a^{3}} \\ & = -(0 +0 + 0) + \frac{1}{4a^3} \\ & = \frac{1}{4a^3} \end{align*} in other case you cand use Gamma function as $$ \int_{0}^{\infty} x^{\alpha-1} e^{-\beta x} = \frac{\Gamma(\alpha)}{\beta^{\alpha}} $$

Answer 2

For such integrals it is useful to find the substitution that will make it similar to Gamma function definition: $$\int_{0}^{+\infty}x^2\exp\{-2ax\}dx = \frac{1}{(2a)^3}\int_{0}^{+\infty}(2ax)^2\exp\{-2ax\} d(2ax) $$ Use the following substitution: $t=-2ax$ if $a > 0$: $$=\frac{1}{(2a)^3}\int_0^{+\infty}t^2\exp\{-t\}dt = \frac{\Gamma(3)}{(2a)^3} = \frac{2}{8a^3} = \frac{1}{4a^3}$$ if $a<0$: $$=\frac{1}{(2a)^3}\int_{0}^{-\infty}t^2\exp\{-t\}dt \to \infty$$ if $a=0$: $$\int_{0}^{+\infty}x^2dx \to \infty$$

Answer 3

Using L’Hospital Rule, we have $$ \lim _{x \rightarrow \infty} x e^{-2 a x}=\lim _{x \rightarrow \infty} \frac{x}{e^{2 a x}}=\lim _{x \rightarrow \infty} \frac{1}{2 a e^{2 a x}}=0 $$ and $$ \lim _{x \rightarrow \infty} x^2 e^{-2 a x}=\lim _{x \rightarrow \infty} \frac{x^2}{e^{2 a x}}=\lim _{x \rightarrow \infty} \frac{2 x}{2 a e^{2 a x}}=0 $$ Therefore via integration by parts twice, we have $$ \begin{aligned}\int_0^{\infty} x^2 e^{-2 a x} d x = & -\frac{1}{2 a} \int_0^{\infty} x^2 d\left(e^{-2 a x}\right) \\ = & -\frac{1}{2 a}\left[x^2 e^{-2 a x}\right]_0^{\infty}+\frac{1}{a} \int_0^{\infty} x e^{-2 a x} d x \\ = & -\frac{1}{2 a^2} \int_0^{\infty} x d\left(e^{-2 a x}\right) \\ = & \frac{1}{2 a^2} \int_0^{\infty} e^{-2 a x} d x \\ = & -\frac{1}{4 a^3}\left[e^{-2 a x}\right]_0^{\infty}\\=&\frac{1}{4 a^3} \end{aligned} $$