[SOLVED] Calculating an improper integral via complex analysis

**arbolis** · October 25th 2009, 10:05 AM

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

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

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

**Chris L T521** · October 25th 2009, 10:34 AM

Originally Posted by arbolis

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

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

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

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

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

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

So it follows then that

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

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

Thus, by the residue theorem,

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

Does this make sense?

**arbolis** · October 25th 2009, 11:18 AM

Originally Posted by Chris L T521

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

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

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

So it follows then that

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

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

Thus, by the residue theorem,

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

Does this make sense?

Yes it does, thank you so much.

**tonio** · October 25th 2009, 11:33 AM

Originally Posted by arbolis

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

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

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 $z^4=-1$ in exponential form:

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

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

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

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

Tonio

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