How Can I Solve This Complex Integral Using Trigonometry or Complex Analysis?

[asi123]

Homework Statement

Hey guys.
I have this integral, I tried to use trigo, tried to use the complex expression but nothing worked, can I please have some help?

Thanks a lot.

Homework Equations

The Attempt at a Solution

 

[gabbagabbahey]

The complex Residues method should work fine; why don't you show us what you tried for that method...

 

[asi123]

The complex Residues method should work fine; why don't you show us what you tried for that method...

Ok, so I tried to use the residues theorem.
I did the substitute but ended up with that really ugly expression, should I use the residues theorem on it?

 

[gabbagabbahey]

I won't be able to read your attempt until admin approves your attachment. You can save time by uploading your file to imageshack.us and posting a link to it instead.

 

[asi123]

I won't be able to read your attempt until admin approves your attachment. You can save time by uploading your file to imageshack.us and posting a link to it instead.

http://img8.imageshack.us/img8/1059/scan0015p.jpg

Thanks a lot.

 

[gabbagabbahey]

Okay, so far so good...now find the 4 roots of the denominator in order to find the poles...which of those poles lies within your contour?

 

[gabbagabbahey]

To make your calculations easier, you should note that if $|z^2|>1$ then so is $|z|$; and $3+2\sqrt{2}>1$.

Also note that $$\sqrt{3-2\sqrt{2}}=\sqrt{2}-1$$

 

[asi123]

To make your calculations easier, you should note that if $|z^2|>1$ then so is $|z|$; and $3+2\sqrt{2}>1$.

Also note that $$\sqrt{3-2\sqrt{2}}=\sqrt{2}-1$$

Oh, I've only now noticed your calculations tips

Anyway, this is what I did:

http://img27.imageshack.us/img27/4732/scan0016y.jpg

I've to warn you, The numbers are awful.

Does it seems right?

Thanks a lot.

 

[gabbagabbahey]

I've to warn you, The numbers are awful.

They sure are!

Luckily, they're also correct...you can simplify them though...after a little algebra you should find that $A=B=\frac{-1}{\sqrt{2}}$ and so your final result becomes $2\pi(\sqrt{2}-1)$

 

[asi123]

They sure are!

Luckily, they're also correct...you can simplify them though...after a little algebra you should find that $A=B=\frac{-1}{\sqrt{2}}$ and so your final result becomes $2\pi(\sqrt{2}-1)$

Thanks a lot