I want to solve the following integral:
$$\int_{-\infty}^\infty k\cdot |Ae^{-a|k-k_0|}|^2dk$$
I did the following:
Substitute $\gamma(k) = k-k_0 \Leftrightarrow k = \gamma + k_0;~\gamma(\pm\infty) = \pm\infty;~d\gamma = dk$ and factoring out constants:
$$\int_{-\infty}^\infty k\cdot |Ae^{-a|k-k_0|}|^2dk = |A|^2\int_{-\infty}^\infty (\gamma + k_0)|e^{-2a|\gamma|}|d\gamma$$
Now one can split the integral and leave out the "||" for the $\exp$ since its always positive:
$$\Rightarrow |A|^2\int_{-\infty}^\infty \gamma e^{-2a|\gamma|}d\gamma + |A|^2k_0\int_{-\infty}^\infty e^{-2a|\gamma|}d\gamma$$
Evaluating the right integral first:
$$|A|^2k_0\int_{-\infty}^\infty e^{-2a|\gamma|}d\gamma = |A|^2k_0\left[\int_{-\infty}^0 e^{2a\gamma} d\gamma + \int_{0}^\infty e^{-2a\gamma} d\gamma\right] $$
$$|A|^2k_0 \left[ \frac{1}{2a} \left. e^{2a\gamma} \right\rvert_{-\infty}^0 - \frac{1}{2a} \left. e^{-2a\gamma} \right\rvert_{0}^\infty \right] = |A|^2k_0 \left[ \frac{1}{2a} + \frac{1}{2a} \right] = \frac{|A|^2k_0}{a}$$
Evaluating the left integral:
$$|A|^2\int_{-\infty}^\infty \gamma e^{-2a|\gamma|}d\gamma = |A|^2\left[ \int_{-\infty}^0\gamma e^{2a\gamma}d\gamma + \int_{0}^\infty \gamma e^{-2a\gamma} d\gamma \right]$$
Using integration by parts choosing $u = \gamma \Rightarrow u' = 1$ and $v' = e^{\pm 2a\gamma} \Rightarrow v = \frac{1}{\pm 2a}e^{\pm 2a\gamma}$ we get:
$$\int_{-\infty}^0\gamma e^{2a\gamma}d\gamma = \left. \frac{1}{2a} \gamma e^{2a\gamma} \right\rvert_{-\infty}^0 - \int_{-\infty}^0 \frac{1}{2a} e^{2a\gamma} d\gamma = -\frac{1}{4a^2} \left. e^{2a\gamma} \right\rvert_{-\infty}^0 = -\frac{1}{4a^2} $$
Analog for the second term this leads to
$$\int_{0}^\infty \gamma e^{-2a\gamma} d\gamma = \frac{1}{4a^2}$$
So that
$$|A|^2\left[ \int_{-\infty}^0\gamma e^{2a\gamma}d\gamma + \int_{0}^\infty \gamma e^{-2a\gamma} d\gamma \right] = |A|^2 \left[-\frac{1}{4a^2} + \frac{1}{4a^2} \right] = 0$$
And thus finally
$$\int_{-\infty}^\infty k\cdot |Ae^{-a|k-k_0|}|^2dk = \frac{|A|^2k_0}{a}$$
I just want to know whether there is some error in my calculations :)