Computing $\int_{0}^{+\infty}\frac{\log(x)}{\sqrt x(1+{x^2})}dx$.

Question

I would like to compute the following integral : $$\int_{0}^{+\infty}\frac{\log(x)}{\sqrt x(1+{x^2})}dx$$ using Residue theorem. I took the contour corresponding to half of the "donuts" $\{r<|z|<R\}$. So by theorem and an classical argument we have $I=2i\pi Res(f,i)$. My problem here, is that I need ton compute the residue for the principal argument of $\log(z)$ and $\sqrt{z}$. How can I do that ?

Note Bene, apologies for the 'unclear' question but I am on my mobile.

Answer 1

Sub $x=y^2$ and the integral becomes

$$4 \int_0^{\infty} dy \frac{\log{y}}{1+y^4} $$

So consider

$$\oint_C dz \frac{\log^2{z}}{1+z^4} $$

where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$ about the positive real axis. In the limit as $R \to \infty$ and $\epsilon \to 0$, the contour integral is equal to

$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{1+x^4} + 4 \pi^2 \int_0^{\infty} \frac{dx}{1+x^4} $$

By the residue theorem, the contour integral is equal to $i 2 \pi$ times the sum of the residues at the poles of the integrand, or

$$i 2 \pi \left[\frac{(i \pi/4)^2}{4 e^{i 3 \pi/4}} +\frac{(i 3\pi/4)^2}{4 e^{i 9 \pi/4}}+\frac{(i 5 \pi/4)^2}{4 e^{i 15 \pi/4}}+\frac{(i 7 \pi/4)^2}{4 e^{i 21 \pi/4}}\right ] = \frac{\sqrt{2} \pi^3}{4} (4+i)$$

Now, by the residue theorem again, we may find similarly that**

$$-i 2 \pi \int_0^{\infty} \frac{dx}{1+x^4} = i 2 \pi \left[\frac{i \pi/4}{4 e^{i 3 \pi/4}} +\frac{i 3\pi/4}{4 e^{i 9 \pi/4}}+\frac{i 5 \pi/4}{4 e^{i 15 \pi/4}}+\frac{i 7 \pi/4}{4 e^{i 21 \pi/4}}\right ] = -i \frac{\sqrt{2} \pi^2}{2}$$

$$\implies \int_0^{\infty} \frac{dx}{1+x^4} = \frac{\sqrt{2} \pi}{4} $$

Putting all of the above results together, we find that

$$ \int_0^{\infty} dx \frac{\log{x}}{1+x^4} = -\frac{\sqrt{2} \pi^2}{16} $$

so that the original integral is

$$\int_0^{\infty} dx \frac{\log{x}}{\sqrt{x}(1+x^2)} = -\frac{\sqrt{2} \pi^2}{4} $$

** I know there are easier ways to get the value of that integral, but I prefer to be consistent with the method I use.

Answer 2

Hint: Let $u=\sqrt{x}$ to avoid the square root and then $x=u^2$ and the integral becomes $$ I=4\int_0^\infty\frac{\log u}{1+u^4}du. $$ Then use the contour $\Gamma=\{z| r\le|z|\le R\}\setminus [r,R]$ to work on the function $$ f(z)=\frac{\log^2 z}{1+z^4} $$ and you will get the anwser.