# Thread: Branch Cuts and Integrals

1. ## Branch Cuts and Integrals

Hello again,

I've come across a new topic that involves solving integrals that involve logarithms. The examples in the text are not too clear how this is done. Can anyone help me solve these:

$\displaystyle \int_{0}^{\infty} \frac{logx}{1+x^4} dx$

$\displaystyle \int_{0}^{\infty} \frac{(logx)^2}{1+x^2} dx$

$\displaystyle \int_{0}^{\infty} \frac{\sqrt{x}logx}{1+x^2} dx$

Any help would be really appreciated. I just can't seem to figure out the right steps to go about solving these problems. I learn backward and need to see examples I understand to get the concept. Thanks in advance!

2. In order to illustrate the procedure we will consider the integral...

$\displaystyle \displaystyle \int_{0}^{\infty} \frac{\ln x}{1+x^{4}}$ (1)

... the others are similar. The integrand function has in $\displaystyle z=0$ a singulatity of the type branch point and we have to select an integration path that doesn't contain this point. The best cadidate is the 'red path' in the figure...

First step is to valuate the integral...

$\displaystyle \displaystyle \int_{c} \frac{\ln z}{1+z^{4}}$ (2)

... where c is the path ABCDEFGHA... how can we proceed?...

Kind regards

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

3. The second step is the computation of the integral...

$\displaystyle \displaystyle \int_{c} \frac{\ln z}{1+z^{4}}$ (1)

... along the 'red path' in the figure. At this scope we use the residue theorem...

$\displaystyle \displaystyle \int_{c} f(z)\ dz = 2\ \pi\ i\ \sum_{k} r_{k}$ (2)

... where...

$\displaystyle \displaystyle r_{k} = \lim_{z \rightarrow z_{k}} (z-z_{k})\ f(z)$ (3)

... are the residues of the poles of f(*) inside the path c. In our case the poles are at $\displaystyle \displaystyle z_{k} = e^{i (2 k+1) \frac{\pi}{4}}$ , $\displaystyle k=0,1,2,3$ so that...

$\displaystyle \displaystyle r_{k} = i\ \frac{\pi}{16}\ (2k+1)\ e^{-i (2k+1) \frac{3}{4} \pi}\implies \int_{c} f(z)\ dz = 2\ \pi\ i\ \sum_{k=0}^{3} r_{k} = -i\ \frac{\pi^{2}}{\sqrt{2}}$ (4)

Honestly in pure calculus I am a little poor [] , so that I propose a little break hoping that somebody verifies my computation, in particular the result (4)...

Kind regards

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

4. I'm afraid that HalsofIvy's congratulations are a little premature for the reason that we will see now. In the last post we have found that...

$\displaystyle \displaystyle \int_{c} \frac{\ln z}{1+z^{4}}\ dz = -i\ \frac{\pi^{2}}{\sqrt{2}}$ (1)

The (1) can be write as...

$\displaystyle \displaystyle \int_{r}^{R} \frac{\ln x}{1+x^{4}}\ dx + i\ R\ \int_{0}^{2 \pi} \frac{\ln R + i\ \theta}{1+R^{4}\ e^{4 i \theta}}\ e^{i \theta}\ d\theta +$

$\displaystyle \displaystyle + \int_{R}^{r} \frac{\ln x + 2\ \pi\ i}{1+x^{4}}\ dx + i\ r\ \int_{2 \pi}^{0} \frac{\ln r + i\ \theta}{1+r^{4}\ e^{4 i \theta}}\ e^{i \theta}\ d\theta = -i\ \frac{\pi^{2}}{\sqrt{2}}$ (2)

Now if in (2) $\displaystyle R \rightarrow \infty$ and $\displaystyle r \rightarrow 0$ the second and the fourth integral vanish and we obtain...

$\displaystyle \displaystyle \int_{0}^{\infty} \frac{\ln x }{1+x^{4}}\ dx - \int_{0}^{\infty} \frac{\ln x }{1+x^{4}}\ dx - 2 \pi i \int_{0}^{\infty} \frac{dx}{1+x^{4}} = -i\ \frac{\pi^{2}}{\sqrt{2}} \implies \int_{0}^{\infty} \frac{dx}{1+x^{4}} = \frac{\pi}{2\ \sqrt{2}}$ (3)

... that is a very good result but... not the goal we were searching for ... may be that we have to take a little different way...

Kind regards

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

5. How about doing the same analysis but use the integral:

$\displaystyle \displaystyle\int \frac{\log^2(z)}{1+z^4}dz$

6. From the 'stubborn old wolf' a new attempt... the function to be integrate is again $\displaystyle f(z) = \frac{\ln z}{1+z^{4}}$ but the new 'red path' is illustrated in figure...

First step is the computation of the residues of the two poles inside the 'red path' and that gives...

$\displaystyle \displaystyle \int_{c} \frac{\ln z}{1+z^{4}}\ dz = 2\ \pi\ i\ \sum_{k=0}^{1} r_{k} = - \frac{\pi^{2}}{4 \sqrt{2}} + i\ \frac{\pi^{2}}{2 \sqrt{2}}$ (1)

As in the previous post coth the contribute of the 'small half circle' for $\displaystyle r \rightarrow 0$ and the contribute of the 'big half circle' for $\displaystyle R \rightarrow \infty$ vanish so that is...

$\displaystyle \displaystyle \int_{c} \frac{\ln z}{1+z^{4}}\ dz = \int_{0}^{\infty} \frac{\ln x}{1+x^{4}}\ dx + \int_{-\infty}^{0} \frac{\ln (-x) }{1+x^{4}}\ dx =$

$\displaystyle \displaystyle = 2\ \int_{0}^{\infty} \frac{\ln x}{1+x^{4}}\ dx - i\ \pi\ \int_{0}^{\infty} \frac{dx}{1+x^{4}} = - \frac{\pi^{2}}{4 \sqrt{2}} + i\ \frac{\pi^{2}}{2 \sqrt{2}}$ (2)

From (2) we derive immediately...

$\displaystyle \displaystyle \int_{0}^{\infty} \frac{\ln x}{1+x^{4}}\ dx = -\frac{\pi^{2}}{8\ \sqrt{2}}$ (3)

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

In Italy we say: due piccioni con una fava! ...

Kind regards

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