Contour integration

1. Apr 28, 2010

Wiseman101

1. The problem statement, all variables and given/known data

By an appropriate choice of contour, validate the following relationship by evaluation,

2. Relevant equations

3. The attempt at a solution
Any help on this one would be much appreciated.

Attached Files:

File size:
2.5 KB
Views:
150
• contour2.png
File size:
3.2 KB
Views:
155
Last edited: Apr 28, 2010
2. Apr 28, 2010

Cyosis

I'm afraid copy pasting the relevant equations don't constitute as a attempt at a solution. What type of contour are you thinking of and do you know how to find the poles?

Edit: I believe the answer should be $\pi/2$

Last edited: Apr 28, 2010
3. Apr 28, 2010

gabbagabbahey

I'd start by finding all the singularities of the integrand...where are they? What type of singularities are they?

4. Apr 29, 2010

Wiseman101

Put in z in place of the x's. The singularities are when the bottom line is zero, so theres a singularity at z = i and z = -i. The contour im thinking would be a semi circle on the positive side of the axis so -i isnt needed so z = i is one simple pole. I dont know how to handle the z^4 though.

5. Apr 29, 2010

Cyosis

That is correct so far. You handle the z^4 term in the same manner, $(z^2+1)(z^4+1)=0$ gives you $(z^2+1)=0$ and $(z^4+1)=0$. Solve for z.

6. Apr 29, 2010

Wiseman101

Ok, so i split ($$z^{2}$$ + 1) into (z+i)(z-i) to get the first pole. So am i right in spliting ($$z^{4}$$ + 1) into ($$z^{2}$$ + i)($$z^{2}$$ - i)? Do i then split each of those again?

7. Apr 29, 2010

Cyosis

Don't need to split them further, just solve for z now.

8. Apr 29, 2010

Wiseman101

Is z = $$\sqrt{i}$$ the other simple pole? I tried to do it out using that but i didnt get out $$\frac{\pi}{6}$$ as the final answer

9. Apr 29, 2010

Cyosis

You should have a total of three poles within the contour, $\sqrt{i}$ is one of them.

As for not getting pi/6, read post #2.

10. Apr 29, 2010

Wiseman101

Got it! Thanks a lot for all your help.