• Support PF! Buy your school textbooks, materials and every day products Here!

Solving a very strange improper integral

  • Thread starter Susanne217
  • Start date
  • #1
317
0

Homework Statement



I am getting fooled by the this improper integral


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

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

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

Answers and Replies

  • #2
1,384
0
is x a constant term here?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Since you are integrating with respect to v, this is just
[tex](cos(x)+ sin(x))\int_0^\infty \frac{dv}{1+ v^2}[/tex]


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

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

Related Threads on Solving a very strange improper integral

  • Last Post
Replies
1
Views
700
  • Last Post
Replies
7
Views
1K
Replies
2
Views
2K
  • Last Post
Replies
3
Views
6K
  • Last Post
Replies
7
Views
794
Replies
4
Views
570
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
1
Views
606
  • Last Post
Replies
3
Views
715
Top