Solving a very strange improper integral

[Susanne217]

Homework Statement

I am getting fooled by the this improper integral


$$\int_0^{\infty}\frac{cos(x)+sin(x)}{1+v^2}dv = \pi \cdot e^{-x}$$

How the devil do I go about getting that result?

The Attempt at a Solution

I end up getting the sum of the two integrals

$$\int_0^{1}\frac{cos(x)}{1+v^2}dv + \int_1^{\infty}\frac{sin(x)}{1+v^2}dv$$ but how do I proceed from there? Hints please :)

 

[zorro]

is x a constant term here?

 

[HallsofIvy]

Since you are integrating with respect to v, this is just
$$(cos(x)+ sin(x))\int_0^\infty \frac{dv}{1+ v^2}$$


And you should know that
$$\int\frac{dv}{1+ v^2}= arctan(v)+ C$$.

But that is NOT equal to $\pi e^{-x}$. You must have copied the problem incorrectly or left out some relationship between x and v.