Residue Theorem #2

• Jul 14th 2009, 09:11 AM
Richmond
Residue Theorem #2
Using residue theorem, calculate

$\int_{-\infty}^\infty \frac {1}{1+x^4} dx$

How do I find the poles of this function?
• Jul 14th 2009, 09:25 AM
pomp
Quote:

Originally Posted by Richmond

$\int_{-\infty}^\infty \frac {1}{1+x^4} dx$

How do I find the poles of this function?

This function will have poles when the denominator is zero.

Just solve $x^4 = -1$
• Jul 14th 2009, 09:00 PM
Richmond
so $x^4 = -1$

$x^2 = \sqrt i$

$x = \pm i^\frac{1}{4}$

Is that right?
• Jul 14th 2009, 09:07 PM
pomp
That is almost right, $x^4 =-1 ~ \Rightarrow x^2 = \pm \sqrt{-1} = \pm i$
What contour would you use to evaluate this integral? Which of the singularities lie within this contour and can you calculate the residues?
• Jul 14th 2009, 09:37 PM
Richmond
The four values are
$i^\frac{1}{4}, -i^\frac{1}{4}, i^\frac{-1}{4},-i^\frac{-1}{4}$

Since the integral limits are $-\infty$ and $\infty$
we take the positive part of the integral and use the positive values

which are
$i^\frac{1}{4}$ and $i^\frac{-1}{4}$

So
$Res(f,i^\frac{1}{4}) = \frac{1}{(x + i^\frac{1}{4})(x + i^\frac{-1}{4}), (x- i^\frac{-1}{4})}$
$= \frac{1}{(2i^\frac{1}{4})(i^\frac{1}{4} + i^\frac{-1}{4})(i^\frac{1}{4} - i^\frac{-1}{4})}$

so the integral will be $2\pi i * Res(f,i^\frac{1}{4})$
=

I think somewhere is wrong.
• Jul 15th 2009, 08:58 AM
HallsofIvy
Quite frankly, the fact that you keep writing " $i^{1/4}$" indicates you need to brush up on "complex numbers" before trying contour integrals and residues.

i can be written in "polar form" as $e^{i\pi/2}$. It's fourth roots are given by $e^{\frac{i(\pi/2+ 2\pi k)}{4}}$ which gives different result for k= 0, 1, 2, and 3.

With k= 0, a fourth root of i is $e^{i\pi/8}= cos(\pi/8)+ i sin(\pi/8)$. With k= 1, another is $e^{5i\pi/8}= cos(5\pi/8)+ i sin(5\pi/8)$. With k= 2, $e^{9i\pi/8}= cos(9\pi/8)+ i sin(9\pi/8)$. Finally, with k= 3, $e^{13\pi/8}= cos(13\pi/8)+ i sin(13\pi/8)$.

Two of those four roots have positive real part, two negative real part. Two have positive imaginary part, two negative imaginary part. That means you really only need to look at the residues at two poles, depending on how you set up the contour.
• Jul 15th 2009, 09:15 AM
Opalg
Quote:

Originally Posted by Richmond
so $x^4 = -1$

$x^2 = \sqrt i$

$x = \pm i^\frac{1}{4}$

Is that right?

No. You want the fourth roots of –1, or in other words the square roots of $\pm i$.
• Jul 15th 2009, 09:25 AM
Random Variable
$x^{4} +1 = 0$

$x^{4} = -1$

$x^{4} = e^{i \pi}$

$x = e^{ \frac{i( \pi + 2 \pi k)}{4}}$ for k = 0, 1, 2, and 3

so the roots are $e^{\frac{i \pi}{4}}, e^{\frac{3 i \pi}{4}}, e^{\frac{5 i \pi}{4}}, e^{\frac{7 i \pi}{4}}$

For the countour integration you only need the two poles in the upper half of complex plane ( $e^{\frac{i \pi}{4}}$ and $e^{\frac{3 i \pi}{4}}$)

$e^{\frac{i \pi}{4}} = \cos(\pi /4) + i \sin(\pi /4) = \sqrt{2}/2+ i \sqrt{2}/2$

$^{\frac{i 3\ pi}{4}} = \cos(3 \pi /4) + i \sin(3 \pi /4) = -\sqrt{2}/2+ i \sqrt{2}/2$
• Jul 15th 2009, 09:51 AM
Random Variable
$\int^{\infty}_{\text{-}\infty} \frac {1}{1+x^{4}}$

let $f(z) = \frac {1}{1+z^4}$

As I mentioned, there are two simple poles in the upper half of the complex plane.

$Res \{f, z_{o}\} = \lim_{z \to z_{0}} (z-z_{0}) \frac {1}{1+z^{4}} = \lim_{z \to z_{0}} \frac {1}{4z^{3}} = \frac {1}{4z_{0}^{3}}$

$Res \{f, \sqrt{2}/2+ i \sqrt{2}/2\} = \text{-} \sqrt{2} /8 - i \sqrt{2} /8$ (after simplification)

$Res \{f, - \sqrt{2}/2+ i \sqrt{2}/2\} = \sqrt{2} /8 - i \sqrt{2} /8$

$\int^{\infty}_{\text{-}\infty} \frac {1}{1+x^{4}} = 2 \pi i \sum (\text{residues in the upper half plane})$ $= 2 \pi i (\text{-}i \sqrt{2} /4 ) = \pi \frac{\sqrt{2}}{2}$