10
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ For one, the result will also depend on $c_1$ and $c_2$. Or are we free to choose these? $\endgroup$
    – Alex M.
    Apr 26, 2019 at 12:29
  • 1
    $\begingroup$ @AlexM., we are free to choose as long as they don't contain $y$. $\endgroup$
    – gbd
    Apr 26, 2019 at 13:15
  • $\begingroup$ This can be done if $c_1= -\infty$ adn $c_2=\infty$. Look into infinite divisibility of Laplace distribution. $\endgroup$
    – Boby
    Apr 26, 2019 at 14:24
  • $\begingroup$ @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. $\endgroup$ Apr 28, 2019 at 20:16

1 Answer 1

5
+50
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .