I do not need a complete proof, just a hint. This is what the problem is:
$$\int_{0}^{\infty} \frac{\log(1+x^2)}{1+x^2} dx$$
Over this contour:
The radius is $R$ from the midpoint.
I am trying to prove somehow:
$$\oint_{A} f(z) dz = 0$$
Using the Ml inequality - Estimation Lemma, ML-inequality.
First we take the first part of the expression:
$$\int_{0}^{\infty} \frac{\log(z+i)}{1 + z^2}$$
The obvious beginning is to parametrize,
$$z = Re^{i\theta}$$
$$\log(Re^{i\theta} + i) < M_1$$
I am first trying to find an upper-bound for $\log(z+i)$
We know:
$$\log(z) = \log|z| + iarg(z)$$
So then, obviously,
$$\log(Re^{i\theta} + i) = \log(|Re^{i\theta}) + i|) + i\theta$$
$$= \log(\sqrt{(Re^{i\theta})^2 - 1}) + i\theta$$
The value of $1$ reduces the logarithm so therefore, And $\theta \le \pi$ so:
$$= \log(\sqrt{(Re^{i\theta})^2 - 1}) + i\theta < \log(R) + 2i\pi \le |\log(R) + 2i\pi| = \sqrt{\log^2(R) - 4\pi^2} \le M_1$$
$$|z^2 + 1| > R^2 - 1$$
$$\therefore, \frac{1}{|z^2 + 1|} < \frac{1}{R^2 - 1}$$
$$M \ge \frac{\sqrt{\log^2(R) - 4\pi^2}}{R^2 - 1}$$
Where $M$ denotes $\max|f(z)|$
$L(A)$ is the arclengthof the contour $A$ simply equal to:
$$L(A) = (1/2)(2)(\pi)(R) = \pi R$$
Hence:
$$\left| \oint_{A} f(z) dz \right | \le ML(A)= \frac{(\pi R)\cdot\sqrt{\log^2(R) - 4\pi^2}}{R^2 - 1}$$
$$\lim_{R \to \infty} \frac{(\pi R)\cdot\sqrt{\log^2(R) - 4\pi^2}}{R^2 - 1} = 0$$
But I am not sure if the $ML$ inequality even applies, because I dont if how to find $\max|f(z)|$
Can someone justify if this work is valid,
Or how to actually use the ML Inequality? Please?
Thanks!
Over $A$, $x=Re^{i\theta}$ with $\theta\in[0,\pi]$. Since: $$ \log(1+x^2) = 2\log(x) + \log\left(1+\frac{1}{x^2}\right),$$ assuming $R\geq e^{\pi}$, for any $x\in A$ we have: $$ \left|\log(1+x^2)\right| \leq 4 \log R, $$ so: $$\left|\oint_{A}\frac{\log(1+x^2)}{1+x^2}\,dx\right|\leq \frac{4\pi R}{R^2-1}\log R \to 0.$$
By the way, there is a clever real analytic technique for evaluating: $$ I = \int_{0}^{+\infty}\frac{\log(1+x^2)}{1+x^2}\,dx=\int_{0}^{\pi/2}\log(1+\tan^2\theta)\,d\theta = -\int_{0}^{\pi}\log\sin t\,dt $$ that is to exploit the identity: $$ \prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{2n}{2^n}.$$ Since, considering Riemann sums: $$\int_{0}^{\pi}\log\sin t\,dt = \lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n},$$ we just have: $$ I = \color{red}{\pi\log 2}.$$