Computing $\mathcal{F}^{-1}\left(\frac{1}{(1+ik)(2+ik)}\right)$ via Convolution Theorem - Heaviside Step Function 
I am trying to solve $$\mathcal{F}^{-1}\left(\frac{1}{(1+ik)(2+ik)}\right),$$ using the convolution theorem. While I am very aware that this problem is more easily solved using partial fractions, this question specifically requires the use of the convolution theorem.

My attempt ($H(x)$ denotes the Heaviside step function):
\begin{align}
\mathcal{F}^{-1}\left(\frac{1}{(1+ik)(2+ik)}\right)&=\sqrt{2\pi}H(x)e^{-x}\ast\sqrt{2\pi}H(x)e^{-2x} \\
&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}2\pi H(x-u)H(u)e^{-x-u} \ du \\
&=\sqrt{2\pi}\int_{-\infty}^{\infty}H(x-u)H(u)e^{-x-u} \ du \tag{1} 
\end{align}
It is at this point I don't know how to proceed. I know that by using partial fractions, the answer is 
$$\mathcal{F}^{-1}\left(\frac{1}{(1+ik)(2+ik)}\right)=\begin{cases} 
      0 & x< 0 \\
      \sqrt{2\pi}(e^{-x}-e^{-2x}) & x>0 \\
   \end{cases}.$$
How do I simplify my integral $(1)$?
 A: Dropping the $\sqrt {2\pi} \ \mathrm e^{-x}$ factor (for my typing convenience), you want to compute
$$\int _{-\infty} ^\infty H(x-u) \ H(u) \ \mathrm e^{-u} \ \mathrm d u$$
where $H(y) = \begin{cases} 0, & y<0 \\ 1, & y \ge 0 \end{cases}$. (There are several conventions for the value of $H$ in $0$, but none of them makes any difference in an integral, because the integral does not "see" individual points, i.e. integrating on $[0,\infty)$ and integrating on $(0,\infty)$ with the Lebesgue measure produce the same result, what happens in $0$ being irrelevant.)
Since $H(u)=0$ for $u<0$, the integral on $(-\infty,0)$ vanishes and your integral is
$$\int _0 ^\infty H(x-u) \ \mathrm e^{-u} \ \mathrm d u \ .$$


*

*If $x<0$ then $x-u < 0$ (because $u \ge 0$), so $H(x-u) = 0$, therefore the above integral is $0$.

*If $x \ge 0$ then
$$\int _0 ^\infty H(x-u) \ \mathrm e^{-u} \ \mathrm d u = \int _0 ^x \underbrace{H(x-u)}_{=1} \ \mathrm e^{-u} \ \mathrm d u + \int _x ^\infty \underbrace{H(x-u)}_{=0} \ \mathrm e^{-u} \ \mathrm d u = \int _0 ^x \mathrm e^{-u} \ \mathrm d u = 1 - \mathrm e^{-x} \ .$$
Putting everything together,
$$\int _{-\infty} ^\infty H(x-u) \ H(u) \ \mathrm e^{-u} \ \mathrm d u = \begin{cases} 0, & x<0 \\ 1 - \mathrm e^{-x}, & x \ge 0 \end{cases}$$
whence, adding the missing $\sqrt {2\pi} \ \mathrm e^{-x}$ factor, you obtain your result.
A: Assume $x>0$ and observe
\begin{align}
e^{-x}\int^\infty_{-\infty}H(x-u)H(u) e^{-u}\ du = e^{-x}\int^x_0 e^{-u}\ du
\end{align}
