calculus - $I =frac{1}{sqrt{2pi}sigma}int_{-infty}^{infty}(1-e^{-kx})cdot e^{-left(frac{(x-z)^2}{2sigma^2}right)} dx.$

Compute $$I
=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}(1-e^{-kx})\cdot e^{-\left(\frac{(x-z)^2}{2\sigma^2}\right)} \ dx,\tag 1$$

where $z$ and $\sigma$ are constants.

I've tried both partial integration to no avail and Im struggling to find a proper substitution.
Are there any apparent tricks to this one?

According to Maple the answer is

$$-\text{csgn}\left(\frac{1}{\sigma}\right)\cdot\Biggl( \exp\left({\frac{k(k\sigma^2-2z)}{2}}\right)-1\Biggr) \tag 2$$

I Googled csgn, and it seems to be a function that is equal to $1$ or $-1$ depending on the sign of $\sigma.$ However, how do I go from $(1)$ to $(2)$?

Answer

We assume $\sigma>0$.

There is a typo in your expression $(2)$, the correct result is
$$
1-e^{\large -kz+\frac{\sigma^2 k^2}{2}}.
$$

One may write
$$\begin{align}
I&=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}(1-e^{-kx})\cdot e^{-\left(\frac{(x-z)^2}{2\sigma^2}\right)} \ dx

\\&=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}e^{-\left(\frac{(x-z)^2}{2\sigma^2}\right)} \ dx-\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}e^{-kx}\cdot e^{-\left(\frac{(x-z)^2}{2\sigma^2}\right)} \ dx
\end{align}
$$ then observe that
$$
-kx-\left(\frac{(x-z)^2}{2\sigma^2}\right)=-\left[\left(\frac{x-z}{\sqrt{2}\,\sigma}+\frac{\sigma k}{\sqrt{2}}\right)^2-\frac{\sigma^2 k^2}{2}+ kz\right]
$$ and one may perform the change of variable
$$
X=\frac{x-z}{\sqrt{2}\,\sigma}+\frac{\sigma k}{\sqrt{2}},\qquad dx=\sqrt{2}\,\sigma \,dX,
$$ in the latter integral giving
$$

I=1-\frac{e^{\large -kz+\frac{\sigma^2 k^2}{2}}}{\sqrt{2\pi}\sigma}\cdot\sqrt{2}\,\sigma \,\int_{-\infty}^{\infty}e^{-X^2} \ dX=1-e^{\large -kz+\frac{\sigma^2 k^2}{2}}
$$ where we have used the gaussian result
$$
\int_{-\infty}^{\infty}e^{-X^2} \ dX=\sqrt{\pi}.
$$ throughout.