9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '19 at 12:29
  • 1
    $\begingroup$ @AlexM., we are free to choose as long as they don't contain $y$. $\endgroup$ – gbd Apr 26 '19 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 '19 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. $\endgroup$ – Isaac Browne Apr 28 '19 at 20:16
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 $y\in \mathbf{R}$, $$ \int_{-\infty}^{\infty} f(y,x) \mathrm d x = \mathrm{e}^{-|y|}.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.