$I =\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}(1-e^{-kx})\cdot e^{-\left(\frac{(x-z)^2}{2\sigma^2}\right)} \ dx.$ 
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 $(2)$?
 A: 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.
A: This integral involves the probability density function of a Normal random variable with mean $z$ and variance $\sigma^2$.  In this context, one always assumes $\sigma>0$.
https://en.wikipedia.org/wiki/Normal_distribution
You can use linearity of the integral to separate the "1" and the "$-e^{-kx}$" into two integrals.
The "1" integral is just an integral over the over the normal probability density function, and is equal to 1, as is any integral over a probability density function.  The proof of this not obvious, and you won't solve it using standard integration techniques for functions of 1 variable.  $e^{-x^2}$ does not have an analytical formula for its antiderivative.
Here is a sketch of the proof:


*

*Consider the square of the integral $\left(\int \frac{1}{\sqrt{2\pi \sigma^2}}\exp\left(-\frac{1}{2\sigma^2}(x-z)^2\right)dx\right)^2$

*Introduce a new dummy variable $y$, and rewrite the square as $\iint \frac{1}{\sqrt{2\pi \sigma^2}}\exp\left(-\frac{1}{2\sigma^2}(x-z)^2\right)\frac{1}{\sqrt{2\pi \sigma^2}}\exp\left(-\frac{1}{2\sigma^2}(y-z)^2\right)dxdy$

*This is now an integral over the 2D $xy$ plane.  Make a change of variables to polar coordinates $(r,\theta)$.  $dxdy$ transforms to $rdrd\theta$ and $(x - z)^2 + (y-z)^2$ in the exponent transforms to $r^2$.  The integrand is constant with respect to $\theta$.

*Now the integral looks like $\int_0^{2\pi}\int_0^\infty \frac{1}{2\pi\sigma^2} \exp(-\frac{1}{2\sigma^2}r^2) rdrd\theta$.  The $\theta$ part just picks up a factor of $2\pi$ since the integral is constant with respect to $\theta$.  Make one more substitution, $t = r^2$ giving $dt = rdr$.  This should give you the result.
The "$-e^{kx}$" integral is the negative of the moment generating function of the normal probability density function.  You can see the result in the same link above.
To prove this you will have to combine the two exponentials, complete the square in the exponent, and then use your previous result.
A: $$I=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty\left(1-e^{-kx}\right)e^{-\frac{(x-z)^2}{2\sigma^2}}dx$$
$$=\frac{1}{\sqrt{2\pi}\sigma}\left(\int_{-\infty}^\infty e^{-\frac{(x-z)^2}{2\sigma^2}}dx-\int_{-\infty}^\infty e^{-kx}e^{-\frac{(x-z)^2}{2\sigma^2}}dx\right)$$
we will first focus on the left integral:
$$I_1=\int_{-\infty}^\infty e^{-\frac{(x-z)^2}{2\sigma^2}}dx$$
first we do a substitution:
$$a=\frac{x-z}{\sqrt{2}\sigma}\,,\frac{da}{dx}=\frac{1}{\sqrt{2}\sigma}$$
so the integral becomes:
$$I_1=\sqrt{2}\sigma\int_{-\infty}^\infty e^{-a^2}da=\sqrt{2\pi}\sigma$$
we have finished $I_1$ so we can now focus on the second integral:
$$I_2=\int_{-\infty}^\infty e^{-kx}e^{-\frac{(x-z)^2}{2\sigma^2}}dx=\int_{-\infty}^\infty e^{-\frac{\left((x-z)^2+2k\sigma^2x\right)}{2\sigma^2}}dx$$
now we must focus on simplifying this exponent:
$$(x-z)^2+2k\sigma^2x=x^2-2xz+z^2+2k\sigma^2x=x^2+2(k\sigma^2-z)+z^2=\left(x+(k\sigma^2-z)\right)^2-(k\sigma^2-z)^2+z^2$$
so the integral now becomes:
$$I_2=\int_{\infty}^\infty e^{-\frac{\left(x+(k\sigma^2-z)\right)^2-(k\sigma^2-z)^2+z^2}{2\sigma^2}}dx=e^{\frac{(k\sigma^2-z)^2-z^2}{2\sigma^2}}\int_{-\infty}^\infty e^{-\left(\frac{x+(k\sigma^2-z)}{\sqrt{2}\sigma}\right)^2}dx$$
now with an easy substitution:
$$u=\frac{x+(k\sigma^2-z)}{\sqrt{2}\sigma}\,,du=\sqrt{2}\sigma du$$
so our integral becomes:
$$I_2=e^{\frac{(k\sigma^2-z)^2-z^2}{2\sigma^2}}\sqrt{2}\sigma\int_{-\infty}^\infty e^{-u^2}du=e^{(k\sigma^2-z)^2-z^2}\sqrt{2\pi}\sigma$$
so now we know that:
$$I=\frac{I_1-I_2}{\sqrt{2\pi}\sigma}=\frac{\sqrt{2\pi}\sigma-\sqrt{2\pi}\sigma e^{\frac{(k\sigma^2-z)^2-z^2}{2\sigma^2}}}{\sqrt{2\pi}\sigma}=1-e^{\frac{(k\sigma^2-z)^2-z^2}{2\sigma^2}}$$
And I believe that is the answer
But this can be simplified to:
$$I=1-e^{\frac{k^2 \sigma^2}{2}-kz}$$
