Integrating the product of the Heaviside function with an exponential

Question

This is a question from my textbook in an applied mathematics class:

On $\mathbb{R}$ with fixed $\alpha >0$, consider the sequences of nonegative continuous functions $$f_k(x) = k^\alpha H(x)xe^{-kx},$$ where $H(x)$ is the Heaviside function.

Part (a) and (b) where to show that $f_k$ converges pointwise (for any $\alpha$) and uniformly (for $\alpha < 1)$ to 0 as $k \rightarrow \infty$, which I have done. Now, in part (c) we have to calculate the integral of the above function from $-\infty$ to $\infty$.

I've entered the formula into Wolfram Alpha to see that the final value is $k^{\alpha-2}$, but unfortunately it doesn't show the actual step-by-step calculation.

Answer 1

For any function $f$ $$ \int_{-\infty}^{+\infty}H(x)f(x)\,dx=\int_0^{+\infty}f(x)\,dx, $$ since $$ H(x)= \begin{cases} 1 & x>0\\ 0 & x<0. \end{cases} $$ Hence, your integral equals $$ k^\alpha\int_0^{+\infty}xe^{-kx}\,dx. $$

Answer 2

Based on mickep's answer, the calculation goes: $$ k^\alpha \int_0^\infty x e^{-kx}dx = k^\alpha \left[\frac{xe^{-kx}}{-k}+ \int_0^\infty \frac{e^{-kx}}{k}dx\right] = k^\alpha \left[\frac{-1}{k^2}e^{-kx} dx\right] = \frac{1}{k^{\alpha-2}}.$$

This appears to be different from the Wolfram Alpha answer, did I make a mistake in this calculation?