1
$\begingroup$

Question:

Find the value of contour integration $$ \int ^{\infty}_0 \frac {\ln x} {(x^2+1)} dx$$

Attempt:

I just calculate

$$\text{Res}(f,z=i) = 2\pi i\lim_{z\to i}(z - i)\frac{\ln z}{z^2+i} = \frac{\pi^2 i}2$$

Im not too sure how to move on from here. Any ideas?

$\endgroup$
  • 1
    $\begingroup$ You need a contour. $\endgroup$ – Fly by Night Apr 28 '16 at 19:24
  • $\begingroup$ I should point out, that if you're not necessarily restriced to complex analysis, this can be done much much more easily. $\endgroup$ – Aritra Das Apr 28 '16 at 20:31
4
$\begingroup$

Consider the branch $f(z) = \frac{\ln z}{z^2 +1}$ where $|z| > 0 , -\frac{\pi}{2}< \arg z < \frac{3\pi}{2}$. Take the path $C = L_2 + L_1 + C_{\rho} + C_R$ where $\rho < 1 < R$ and $C_R$ and $C_{\rho}$ are the semi-circles with radius $R$ and $\rho$ respectively. See the figure below.

$\hskip.75in$enter image description here

By Cauchy's Theorem we have

$$\int_{L_1} f(z ) dz + \int_{L_2} f(z) \, dz + \int_{C_\rho} f(z) \, dz + \int_{C_R} f(z) \, dz = 2\pi i \,\,\mathrm {Res}_{z = i} f(z)$$

Now if $z = re^{i\theta}$ then we may write

$$f(z) = \frac{\ln r + i\theta}{r^2e^{2i\theta} + 1}$$

and use the parametric representations

$$z = re^{i0} = r \,\,(\rho \leq r \leq R) \,\,\, \text{and}\,\,\,z = re^{i\pi} \,\,(\rho \leq r \leq R)$$

for the legs $L_1$ and $-L_2$ respectively, which yields

$$\int_{L_1} f(z)\, dz - \int_{-L_2} f(z) \, dz = \int_{\rho}^R \frac{\ln r}{r^2 +1} \, dr + \int_{\rho}^R \frac{\ln r + i\pi}{r^2 +1} \, dr$$

The residue at $z = i$ is $\mathrm {Res}_{z=i} = \frac{\pi}{4}$, then

$$\int_{\rho}^R \frac{\ln r}{r^2 +1} \, dr + \int_{\rho}^R \frac{\ln r + i\pi}{r^2 +1} \, dr = \frac{\pi^2i}{2} -\int_{C_\rho} f(z) \, dz - \int_{C_R} f(z) \, dz $$

equating the real parts

$$2 \int_{\rho}^R \frac{\ln r}{r^2 +1} \, dr = -\int_{C_\rho} f(z) \, dz - \int_{C_R} f(z) \, dz $$

It remains only to show that $\displaystyle \lim_{\rho \to 0} \int_{C_\rho} f(z) \, dz = 0 $ and $\displaystyle \lim_{R\to \infty} \int_{C_R} f(z) \, dz = 0$.

Do you think you can take it from here?

$\endgroup$
  • $\begingroup$ I'm sorry but what are $L_1, L_2$ ? $\endgroup$ – Aritra Das Apr 28 '16 at 20:07
  • 1
    $\begingroup$ The legs connecting the two semi-circles. It'd be better if I had a drawing made. Working on that. $\endgroup$ – Aaron Maroja Apr 28 '16 at 20:09
  • 1
    $\begingroup$ Take a look at mine. Just finished it. $\endgroup$ – Aaron Maroja Apr 28 '16 at 20:48
  • 1
    $\begingroup$ Very nice, what software did you use? $\endgroup$ – Aritra Das Apr 28 '16 at 20:49
  • 2
    $\begingroup$ Corel Draw. I still can't use tikz on latex, I'm learning it yet. $\endgroup$ – Aaron Maroja Apr 28 '16 at 20:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.