Integral $\int_0^∞ x^2 \exp(-2kx)\,dx$ not defined when putting limits

Question

I have this integral : $\displaystyle\int_0^∞ x^2 \exp(-2kx)\,dx$

now, integrating by parts ($x^2$ as first function and $\exp(-2kx)$ as second),

\begin{align} I & = \int_0^∞ x^2 \exp(-2kx)\,dx \\[6pt] & = \frac{x^2 \exp (-2kx)}{-2k} - \int_0^∞ \frac{2x \exp(-2kx)}{-2k} \,dx \\[6pt] & = \frac{x^2 \exp (-2kx)}{-2k} + \left[ \frac{x \exp(-2kx)}{-2k} - \int_0^∞\frac{\exp (-2kx)}{-2k}\,dx\right] \\[6pt] & = \left.\frac{x^2 \exp (-2kx)}{-2k} + \frac{x \exp(-2kx)}{-2k} + \frac{\exp (-2kx)}{-2k} \right|_0^∞ \\[6pt] & = \left.\frac{\exp(-2kx) (x^2 + x + 1)}{-2k} \right|_0^∞ \end{align}

now how to solve this? have i done the integral right? suggest some other ways if possible.

Answer 1

Assuming $k\in\mathbb{R}^+$ we have:

$$ I_k = \int_{0}^{+\infty} x^2 e^{-2k x}\,dx = \frac{1}{(2k)^3}\int_{0}^{+\infty} z^2 e^{-z}\,dz = \frac{2}{8k^3} = \color{red}{\frac{1}{4k^3}}.$$

The integral appearing in the middle term is just the integral defining a value of the $\Gamma$ function :

$$ \forall n\in\mathbb{N},\qquad \int_{0}^{+\infty} x^n e^{-x}\,dx = n!.$$

Answer 2

What we usually mean whenever we write $$ \int^{\infty}_0 x^2 e^{-2kx}\, dx $$ is $$ \lim_{M\to \infty} \int^{M}_0 x^2 e^{-2kx}\, dx. $$ Thus, we compute the integral (from $0$ to $M$) using the integration by parts formula twice: \begin{align*} \int_0^M x^2e^{-2kx}\, dx & = \Bigg[-\frac{x^2e^{-2kx}}{2k}\Bigg]^M_0 +\int_0^Mx\frac{e^{-2kx}}{k}\, dx = -\frac{M^2e^{-2Mk}}{2k} + \Bigg[-x\frac{e^{-2kx}}{2k^2}\Bigg]^M_0 \\ & +\int_0^M \frac{e^{-2kx}}{2k^2}\, dx = -\frac{e^{-2Mk}}{2k}\Big(M^2+\frac{M}{k}\Big) + \Bigg[-\frac{e^{-2kx}}{4k^3}\Bigg]^M_0 \\ & = -\frac{e^{-2Mk}}{2k}\Big(M^2+\frac{M}{k}\Big) - \frac{e^{-2Mk}}{4k^3} + \frac{1}{4k^3}. \end{align*}

Then, for any $k>0$, we have $$ \int^{\infty}_0 x^2 e^{-2kx}\, dx= \lim_{M\to\infty} -\frac{e^{-2Mk}}{2k}\Big(M^2+\frac{M}{k}\Big) - \frac{e^{-2Mk}}{4k^3} + \frac{1}{4k^3} = \frac{1}{4k^3}. $$

Answer 3

Hint: Alternately, evaluate $I(k)=\displaystyle\int_0^\infty e^{-2kx}~dx$, then differentiate both sides twice with regard to the parameter k.