# Is there a way to preform this integral such that the answer is $e^{-|y|}$?

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.

• For one, the result will also depend on $c_1$ and $c_2$. Or are we free to choose these? – Alex M. Apr 26 '19 at 12:29
• @AlexM., we are free to choose as long as they don't contain $y$. – gbd Apr 26 '19 at 13:15
• This can be done if $c_1= -\infty$ adn $c_2=\infty$. Look into infinite divisibility of Laplace distribution. – Boby Apr 26 '19 at 14:24
• @Boby Can you explain a bit more? I looked into infinite divisibility and do not really see how it implies a solution to this problem. – Isaac Browne Apr 28 '19 at 20:16

## 1 Answer

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 $$y\in \mathbf{R}$$, $$\int_{-\infty}^{\infty} f(y,x) \mathrm d x = \mathrm{e}^{-|y|}.$$