TITLE: Is there a way to preform this integral such that the answer is $e^{-|y|}$?
QUESTION [9 upvotes]: Consider the function $f(y)=e^{-|y|}e^{y}$
I am trying to integrate this function with respect to another variable (such as $x$) so that the result from the integration is $e^{-|y|}$?
The function $f(y)$ can be changed in anyway as long as 
1) the powers of $y$ and $|y|$ stay equal to one. 
2) and the boundaries of integration do not include $y$ in them.
3) The result of the integral is $e^{-|y|}$. Of course the answer may be of the form $A e^{-a|y+b|}$ where $A$, $a$ and $b$ are constants.
So for example we can add a constant or $x$ inside the $||$ or multiply $y$.
So for example the integral
$$\int^{c_2}_{c1}e^{-|x y+a|}e^{y/x}dx$$
or
$$\int^{c_2}_{c1}(e^{-|ay+x|}e^{y+b}+d)dx$$
Is there a way to integrate this so that the answer is  $e^{-|y|}$?
We are free to choose where to put the constants or $x$ as long as the three conditions are satisfied.

REPLY [5 votes]: Here is my attempt, it builds on Cauchy's integral theorem and requires complex numbers, but it works :)
Let
$$ f(y,x) = \frac{1}{\pi}\frac{\mathrm{e}^{-\frac{|y|}{2}}\mathrm e^{y \frac{i x}{2}}}{1 + x^2},$$
Then, for $y\in \mathbf{R}$,
$$ \int_{-\infty}^{\infty} f(y,x) \mathrm d x = \mathrm{e}^{-|y|}.$$