Integral using residue theorem

[shebbbbo]

The question asks to show using the residue theorem that

\intcos(x) / (x² +1)² dx = \pi / e

(the terminals of the integral are -∞ to ∞ but i didnt know the code to write that)

I found the singularities at -i and +i

so i think we change the function inside the integral to cos(z) / (z² +1)²

i expanded the cos(z) as cosh(1) - isinh(1)(z-i) -0.5cosh(1)(z-i)² +...

and i expanded (z²+1)² as -(1/4)(z-i)² - i/4(z-i) + 3/16 +...

I did the same for the singularity at x=-i and when i added both the residues together i got

(9/16e + e/16) (this is multiplied by 2\pii to find residues)

this doesn't seem right? i don't know if what I've done is the right method. please help, I've spent soooo many hours on this one stupid question :(

 

[lanedance]

why do you add singularities together? not too sure what you're attempting...

i think the idea is you need to set up a closed contour in complex space and integrate around it. Then teh vlaue of teh integral will be related to singularities contained within the contour.

a good one would be along the x-axis for -R to R, and a semi-circle in the upper half of the complex plane to connect the contour. Then take the limit r->infinity. Hopefully the semi-cirlce contribution tends to zero, then you can relate the x-axis component directly to the original integral

 

[HallsofIvy]

The question asks to show using the residue theorem that

\intcos(x) / (x² +1)² dx = \pi / e

(the terminals of the integral are -∞ to ∞ but i didnt know the code to write that)

I found the singularities at -i and +i

so i think we change the function inside the integral to cos(z) / (z² +1)²

Wouldn't you have to do that first, in order to find the singularities?

i expanded the cos(z) as cosh(1) - isinh(1)(z-i) -0.5cosh(1)(z-i)² +...

and i expanded (z²+1)² as -(1/4)(z-i)² - i/4(z-i) + 3/16 +...

I did the same for the singularity at x=-i and when i added both the residues together i got

(9/16e + e/16) (this is multiplied by 2\pii to find residues)

this doesn't seem right? i don't know if what I've done is the right method. please help, I've spent soooo many hours on this one stupid question :(

What path are you integrating over? Does it include both singularities? Obviously to do this real integral as a path integral, you need a closed path that includes the real axis. I don't see how you can do that and get both i and -i inside the path.

 

[shebbbbo]

oh yeah... that makes sense...

your right we can't get i and -i in on a path integral across the real axis

maybe i need to build a circle or something? i guess i have less of an idea now?

thanks for your help?

any idea where i should go from here?