# [SOLVED] Calculating an improper integral via complex analysis

• Oct 25th 2009, 10:05 AM
arbolis
[SOLVED] Calculating an improper integral via complex analysis
I must calculate $\displaystyle \int _0 ^{\infty} \frac{dx}{x^4+1}$.
My attempt :
Let $\displaystyle f(z)=\frac{1}{z^4+1}$. I've calculated the poles of $\displaystyle f$ to be $\displaystyle z_i=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$, $\displaystyle \frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$, $\displaystyle -\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$ and $\displaystyle -\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$.

I chose the curve $\displaystyle \gamma _R$ as being the circle centered at $\displaystyle z=0$ with radius $\displaystyle R>1$.
So I have that $\displaystyle \int _{\gamma _R} f(z)dz=2 \pi i \sum _i Res(f,z_i)$.
It remains to calculate the residues of $\displaystyle f$, which I wasn't able to do.
I tried to express $\displaystyle f$ as a Laurent series but I didn't succeed in it.
I also tried to use the fact that the residues are worth $\displaystyle \lim _{z \to z_i} (z-z_i)f(z)$ but without success in evaluating the limit.
So how could I find the residues of $\displaystyle f$?

Thanks in advance... and sorry! I find all this very difficult to grasp in such a few time.
• Oct 25th 2009, 10:34 AM
Chris L T521
Quote:

Originally Posted by arbolis
I must calculate $\displaystyle \int _0 ^{\infty} \frac{dx}{x^4+1}$.
My attempt :
Let $\displaystyle f(z)=\frac{1}{z^4+1}$. I've calculated the poles of $\displaystyle f$ to be $\displaystyle z_i=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$, $\displaystyle \frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$, $\displaystyle -\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$ and $\displaystyle -\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$.

I chose the curve $\displaystyle \gamma _R$ as being the circle centered at $\displaystyle z=0$ with radius $\displaystyle R>1$.
So I have that $\displaystyle \int _{\gamma _R} f(z)dz=2 \pi i \sum _i Res(f,z_i)$.
It remains to calculate the residues of $\displaystyle f$, which I wasn't able to do.
I tried to express $\displaystyle f$ as a Laurent series but I didn't succeed in it.
I also tried to use the fact that the residues are worth $\displaystyle \lim _{z \to z_i} (z-z_i)f(z)$ but without success in evaluating the limit.
So how could I find the residues of $\displaystyle f$?

Thanks in advance... and sorry! I find all this very difficult to grasp in such a few time.

First note that $\displaystyle \int_0^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\int_ {-\infty}^{\infty}\frac{\,dx}{x^4+1}$

Let $\displaystyle z_1=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$, $\displaystyle z_2=\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$, $\displaystyle z_3-\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$ and $\displaystyle z_4=-\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$

If we select our closed contour to be the semicircle above the real axis, we see that we will have residues at $\displaystyle z_1$ and $\displaystyle z_3$.

So it follows then that

$\displaystyle \text{Res}\!\left(f,z_1\right)=\lim_{z\to z_1}\left(z-z_1\right)\frac{1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)}$ $\displaystyle =\lim_{z\to z_1}\frac{1}{(z-z_2)(z-z_3)(z-z_4)}=\frac{1}{(i\sqrt{2})(\sqrt{2})(\sqrt{2}(1+i) )}=\frac{i+1}{-4\sqrt{2}}$

$\displaystyle \text{Res}\!\left(f,z_3\right)=\lim_{z\to z_3}\left(z-z_3\right)\frac{1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)}$ $\displaystyle =\lim_{z\to z_3}\frac{1}{(z-z_1)(z-z_2)(z-z_4)}=\frac{1}{(-\sqrt{2})(\sqrt{2}(i-1))(i\sqrt{2})}=\frac{i-1}{-4\sqrt{2}}$

Thus, by the residue theorem,

$\displaystyle \int_{0}^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\in t_{-\infty}^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\lef t(2\pi i \left(\frac{i+1}{-4\sqrt{2}}+\frac{i-1}{-4\sqrt{2}}\right)\right)$ $\displaystyle =\frac{\pi}{4}\sqrt{2}$

Does this make sense?
• Oct 25th 2009, 11:18 AM
arbolis
Quote:

Originally Posted by Chris L T521
First note that $\displaystyle \int_0^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\int_ {-\infty}^{\infty}\frac{\,dx}{x^4+1}$

Let $\displaystyle z_1=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$, $\displaystyle z_2=\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$, $\displaystyle z_3-\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$ and $\displaystyle z_4=-\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$

If we select our closed contour to be the semicircle above the real axis, we see that we will have residues at $\displaystyle z_1$ and $\displaystyle z_3$.

So it follows then that

$\displaystyle \text{Res}\!\left(f,z_1\right)=\lim_{z\to z_1}\left(z-z_1\right)\frac{1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)}$ $\displaystyle =\lim_{z\to z_1}\frac{1}{(z-z_2)(z-z_3)(z-z_4)}=\frac{1}{(i\sqrt{2})(\sqrt{2})(\sqrt{2}(1+i) )}=\frac{i+1}{-4\sqrt{2}}$

$\displaystyle \text{Res}\!\left(f,z_3\right)=\lim_{z\to z_3}\left(z-z_3\right)\frac{1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)}$ $\displaystyle =\lim_{z\to z_3}\frac{1}{(z-z_1)(z-z_2)(z-z_4)}=\frac{1}{(-\sqrt{2})(\sqrt{2}(i-1))(i\sqrt{2})}=\frac{i-1}{-4\sqrt{2}}$

Thus, by the residue theorem,

$\displaystyle \int_{0}^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\in t_{-\infty}^{\infty}\frac{\,dx}{x^4+1}=\frac{1}{2}\lef t(2\pi i \left(\frac{i+1}{-4\sqrt{2}}+\frac{i-1}{-4\sqrt{2}}\right)\right)$ $\displaystyle =\frac{\pi}{4}\sqrt{2}$

Does this make sense?

Yes it does, thank you so much. :D(Bow)(Handshake)
• Oct 25th 2009, 11:33 AM
tonio
Quote:

Originally Posted by arbolis
I must calculate $\displaystyle \int _0 ^{\infty} \frac{dx}{x^4+1}$.
My attempt :
Let $\displaystyle f(z)=\frac{1}{z^4+1}$. I've calculated the poles of $\displaystyle f$ to be $\displaystyle z_i=\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$, $\displaystyle \frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$, $\displaystyle -\frac{\sqrt 2}{2}+i\frac{\sqrt2}{2}$ and $\displaystyle -\frac{\sqrt 2}{2}-i\frac{\sqrt2}{2}$.

I chose the curve $\displaystyle \gamma _R$ as being the circle centered at $\displaystyle z=0$ with radius $\displaystyle R>1$.
So I have that $\displaystyle \int _{\gamma _R} f(z)dz=2 \pi i \sum _i Res(f,z_i)$.
It remains to calculate the residues of $\displaystyle f$, which I wasn't able to do.
I tried to express $\displaystyle f$ as a Laurent series but I didn't succeed in it.
I also tried to use the fact that the residues are worth $\displaystyle \lim _{z \to z_i} (z-z_i)f(z)$ but without success in evaluating the limit.
So how could I find the residues of $\displaystyle f$?

Thanks in advance... and sorry! I find all this very difficult to grasp in such a few time.

i think that it'll be easier to work if you write the roots of $\displaystyle z^4=-1$ in exponential form:

$\displaystyle z_i=e^{\frac{\pi i}{4}(1+2k)}\,,\,\,k=0,1,2,3$ (instead exponential you can use cis, of course)

Now, write $\displaystyle z^4+1=\prod \limits_{i=1}^{4}(z-z_i)$

And now the evaluation of the limits for the residues is easier, imo. For instance, the residue at $\displaystyle z_2=e^{\frac{5\pi i}{4}}=-\frac{1}{\sqrt{2}}(1+i)$ we get:

$\displaystyle \lim_{z \rightarrow z_2}(z-z_2)f(z)=\lim_{z \rightarrow z_2}\frac{1}{(z-z_0)(z-z_1)(z-z_3)}=\frac{1}{(z_2-z_0)(z_2-z_1)(z_2-z_3)}$ $\displaystyle =\frac{1}{\frac{1}{2\sqrt{2}}(-2(1+i)(-2i)(-2)}=\frac{1+i}{4\sqrt{2}}$ (check this very carefully because there's a good chance I got it wrong)

Tonio