[SOLVED] Solving an integral via complex analysis

**arbolis** · October 18th 2009, 01:00 PM

I must prove that $\int _0^{\infty} \frac{x^2dx}{(x^2+1)(x^2+4)}=\frac{\pi}{6}$ .

My attempt : None, I don't know where to start. I'm really lost, I don't have any textbook but my classnotes.
I'd like a push in the right direction rather than a full answer, but everything's welcome.

**chisigma** · October 18th 2009, 01:47 PM

A possible solution is to integrate, using the residue theorem, the function...

$f(z)= \frac{z^{2}}{(z^{2}+1)(z^{2}+4)}$

... along the grey path of the figure...

... and let R tend to infinity. By Jordan's lemma is...

$\lim_{R \rightarrow \infty} \int_{ABC} f(z)\cdot dz=0$

... and the singularities of $f(*)$ inside the path are $z=i$ and $z=2i$ . Since $f(x)$ is an even function respect to x it will be...

$\int_{0}^{\infty} f(x)\cdot dx = \frac{1}{2} \int_{-\infty}^{+\infty} f(x)\cdot dx$

Kind regards

$\chi$ $\sigma$

**arbolis** · October 18th 2009, 02:39 PM

Originally Posted by chisigma

A possible solution is to integrate, using the residue theorem, the function...

$f(z)= \frac{z^{2}}{(z^{2}+1)(z^{2}+4)}$

... along the grey path of the figure...

... and let R tend to infinity. By Jordan's lemma is...

$\lim_{R \rightarrow \infty} \int_{ABC} f(z)\cdot dz=0$

... and the singularities of $f(*)$ inside the path are $z=i$ and $z=2i$ . Since $f(x)$ is an even function respect to x it will be...

$\int_{0}^{\infty} f(x)\cdot dx = \frac{1}{2} \int_{-\infty}^{+\infty} f(x)\cdot dx$

Kind regards

$\chi$ $\sigma$

Moltissimo grazie, ma... why do you write $f(*)$ and not $f(z)$ ?

Also, is $\lim_{R \rightarrow \infty} \int_{ABC} f(z)\cdot dz=0$ equal to $\lim_{R \rightarrow \infty} \int_{-R}^R f(z)\cdot dz$ ?
I don't think so, as there is 2 singularities as you pointed out. I think I'm misunderstanding something.
So I should calculate $\int _ {\gamma _1} f(z) dz$ where $\gamma_1$ is a closed path around $i$ ? Then I sum this result to $\int _ {\gamma _2} f(z) dz$ where $\gamma _2$ is a closed path around $2i$ . Unless I'm not getting it.

**chisigma** · October 18th 2009, 10:41 PM

The residue theorem states that, if C is a closed path and a function f(*) has a finite number n of simple poles inside C, is...

$\int_{C} f(z) = 2\pi i \sum_{i=1}^{n} r_{i}$ (1)

... where...

$r_{i} = \lim_{z \rightarrow z_{i}} (z-z_{i}) f(z)$ (2)

In our example is...

$f(z)= \frac{z^{2}}{(z^{2}+1)(z^{2}+4)}$ (3)

... C is the gray path in figure...

... for the Jordan's lermma is...

$\lim_{R \rightarrow \infty} \int_{ABC} f(z)\cdot dz= 0$ (4)

... so that, combining (1),(2),(3) and (4), is...

$\int_{-\infty}^{\infty}\frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)}= 2\pi i\cdot \sum_{i=1}^{n} r_{i}$ (5)

In our case f(*) has only two poles inside C : $z_{1}=i$ , $z_{2}=2i$ , so that is...

$r_{1}= \lim_{z \rightarrow i} (z-i)\cdot f(z)= \frac{i}{6}$

$r_{2}= \lim_{z \rightarrow 2i} (z-2i)\cdot f(z)= -\frac{i}{3}$ (6)

By (1) and (5) we have finally...

$\int_{-\infty}^{\infty} \frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)} = \frac{\pi}{3}$ (7)

... so that...

$\int_{0}^{\infty} \frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)} = \frac{\pi}{6}$ (8)

Kind regards

$\chi$ $\sigma$

**arbolis** · October 19th 2009, 07:07 AM

Originally Posted by chisigma

The residue theorem states that, if C is a closed path and a function f(*) has a finite number n of simple poles inside C, is...

$\int_{C} f(z) = 2\pi i \sum_{i=1}^{n} r_{i}$ (1)

... where...

$r_{i} = \lim_{z \rightarrow z_{i}} (z-z_{i}) f(z)$ (2)

In our example is...

$f(z)= \frac{z^{2}}{(z^{2}+1)(z^{2}+4)}$ (3)

... C is the gray path in figure...

... for the Jordan's lermma is...

$\lim_{R \rightarrow \infty} \int_{ABC} f(z)\cdot dz= 0$ (4)

... so that, combining (1),(2),(3) and (4), is...

$\int_{-\infty}^{\infty}\frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)}= 2\pi i\cdot \sum_{i=1}^{n} r_{i}$ (5)

In our case f(*) has only two poles inside C : $z_{1}=i$ , $z_{2}=2i$ , so that is...

$r_{1}= \lim_{z \rightarrow i} (z-i)\cdot f(z)= \frac{i}{6}$

$r_{2}= \lim_{z \rightarrow 2i} (z-2i)\cdot f(z)= -\frac{i}{3}$ (6)

By (1) and (5) we have finally...

$\int_{-\infty}^{\infty} \frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)} = \frac{\pi}{3}$ (7)

... so that...

$\int_{0}^{\infty} \frac{x^{2}\cdot dx}{(x^{2}+1)(x^{2}+4)} = \frac{\pi}{6}$ (8)

Kind regards

$\chi$ $\sigma$

Moltissimo grazie di nuovo!
Very well explained.
If you don't mind me asking a last question : Are these type of integrals common in physics?

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