Integral Using Residue Theorem?

[yeahhyeahyeah]

Homework Statement

the integral of 1/(1+x^4) from -infinity to +infinity

Homework Equations

Residue theorem.

The Attempt at a Solution

1/(1+z^4) so z^4 = -1

I know I should be using the residues at z = -sqrt(i) and z= i*sqrt(i)

I am getting a complex number as an answer which makes no sense

residue at z = -sqrt(i) = 1/(4*i*sqrt(i))
and at z = i*sqrt(i) = 1/(4*sqrt(i))

and therefore integral of (1/1+z^4) = 2pi*i* sum of those residues


Am I on the right track?

 

[Winzer]

This is a tricky. When else does $$z^4+1 =0$$? Maybe at $$z=e^{i\frac{\pi}{4}}$$.
Can you get it from here?

 

[lanedance]

been a while since I've done these, but as a start, rather than working with sqrt i, which i don't think is unique, I would find the singularities by first letting
$$z = r e^{j \theta}$$

whilst for some integer n
$$-1 = e^{j\pi(2n+1)}$$

then the singularities can be found by
$$z^4 = r^4 e^{j 4 \theta} = e^{j\pi(2n+1)}$$

giving
$$\theta = \frac{\pi(2n+1)}{4} = \pi(n/2+1/4)$$

which I think will give 4 unique singularities for n = 0, 1, 2, 3

 

[yeahhyeahyeah]

OOOh, I see, I used the singularities and I get pi/sqrt2 which I think is right. Thank you guys so much. I actually have another related question now though. First of all, why does sqrt(i) not work when I use it as z?

Also,

I'm now trying to do:

integral (0 to infinity) of sin(x^2) using residue theorem as well. I can't figure out how to make the substitution.


I don't know when/if you guys will answer that question but thanks so much for your help just now!

 

[lanedance]

well first you didn't find all the singularities (2 out of 4) & sqrt(i) is not unique. There are 2 unique answers, note we try and solve
$$z^2 = i$$

so as before
$$(e^{i \theta})^2 = e^{i 2 \theta} = e^{(2n+1)\pi}$$

so the 2 solutions in the \theta range [0, 2pi) are
$$\theta = (n+1/2)\pi} = \pi/2, 3\pi/2$$

compare it with solving
$$z^2 = 1$$
there is a plus & minus solution

that said if i remember residues correctly, you probably only need 2 of them anyway

not too sure about your sin question, but it may be worth trying writing the sin in terms of the sum/differnce of 2 complex exponentials

 

[Winzer]

been a while since I've done these, but as a start, rather than working with sqrt i, which i don't think is unique, I would find the singularities by first letting
$$z = r e^{j \theta}$$

whilst for some integer n
$$-1 = e^{j\pi(2n+1)}$$

then the singularities can be found by
$$z^4 = r^4 e^{j 4 \theta} = e^{j\pi(2n+1)}$$

giving
$$\theta = \frac{\pi(2n+1)}{4} = \pi(n/2+1/4)$$

which I think will give 4 unique singularities for n = 0, 1, 2, 3

I agree. Except when you make the semicircle, the poles n=0, 1 are the only ones on the upper half plane.