# Thread: [SOLVED] Solving an integral via complex analysis

1. ## [SOLVED] Solving an integral via complex analysis

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

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.

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

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by chisigma
A possible solution is to integrate, using the residue theorem, the function...

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

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

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

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

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

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

Kind regards

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

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

4. 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...

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

... where...

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

In our example is...

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

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

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

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

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

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

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

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

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

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

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

... so that...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. 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...

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

... where...

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

In our example is...

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

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

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

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

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

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

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

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

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

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

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

... so that...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
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?