# Branch Cuts and Integrals

• Nov 14th 2010, 04:50 PM
KatyCar
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:

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

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

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

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!
• Nov 15th 2010, 08:44 PM
chisigma
In order to illustrate the procedure we will consider the integral...

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

... the others are similar. The integrand function has in $z=0$ 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...

http://digilander.libero.it/luposabatini/MHF85.bmp

First step is to valuate the integral...

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

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

Kind regards

$\chi$ $\sigma$
• Nov 16th 2010, 06:03 AM
chisigma
http://digilander.libero.it/luposabatini/MHF85.bmp

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

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

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

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

... where...

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

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

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

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

Kind regards

$\chi$ $\sigma$
• Nov 16th 2010, 12:47 PM
chisigma
http://digilander.libero.it/luposabatini/MHF85.bmp

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...

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

The (1) can be write as...

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

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

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

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

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

Kind regards

$\chi$ $\sigma$
• Nov 16th 2010, 03:29 PM
shawsend
How about doing the same analysis but use the integral:

$\displaystyle\int \frac{\log^2(z)}{1+z^4}dz$
• Nov 17th 2010, 12:54 PM
chisigma
From the 'stubborn old wolf' a new attempt... the function to be integrate is again $f(z) = \frac{\ln z}{1+z^{4}}$ but the new 'red path' is illustrated in figure...

http://digilander.libero.it/luposabatini/MHF90.bmp

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

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

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

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

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

From (2) we derive immediately...

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

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

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

Kind regards

$\chi$ $\sigma$