# Complex Integral : How can I evaluate $\int_{C_R} f(z) dz$?

I have to calculate $$I$$ ,using complex integral. $$$$I:=\displaystyle\int_{-\infty}^{\infty} \dfrac{\text{log}{\sqrt{x^2+a^2}}}{1+x^2}.(a>0)$$$$

Let $$f(z)=\dfrac{\text{log}(z+ia)}{1+z^2}$$.

$$C_R : z=Re^{i\theta}, \theta : 0 \to \pi.$$

$$C_1 : z=t, t : -R \to R.$$

From Residue Theorem, $$$$\displaystyle\int_{C_1} f(z) dz + \displaystyle\int_{C_R} f(z) dz = 2\pi i \text{Res}(f, i).$$$$

If $$R \to \infty$$, $$\displaystyle\int_{C_1} f(z) dz \to \displaystyle\int_{-\infty}^{\infty} f(x) dx.$$

And my mathematics book says that if $$R \to \infty$$, $$\displaystyle\int_{C_R} f(z) dz \to 0.$$ I cannot understand why this holds.

My attempt is following :

\begin{align} \Bigg|\displaystyle\int_{C_R} f(z) dz \Bigg| &=\Bigg|\displaystyle\int_0^{\pi} f(Re^{i\theta}) i Re^{i\theta} d\theta \Bigg| \\ &\leqq \displaystyle\int_0^{\pi} \Bigg| R \dfrac{\text{log} (Re^{i\theta} + ia)}{1+R^2e^{2i\theta}} \Bigg| d\theta \\ &\leqq \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (Re^{i\theta}+ia) \Bigg| d\theta \\ &= \displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (R\cos \theta+i(R\sin \theta +a)) \Bigg| d\theta \end{align}

I expect that $$\displaystyle\int_0^{\pi} \dfrac{R}{R^2-1} \Bigg| \text{log} (R\cos \theta+i(R\sin \theta +a)) \Bigg| d\theta \to 0$$, but I cannot prove this.

I would like to give me some ideas.

• Maybe use $\log (A+iB)=\frac{1}{2}\log (A^2+B^2) +i\phi$ where $\phi$ is of course small to get a good bound on $|\log (R\cos \theta+i(R\sin \theta +a))|$? – ancientmathematician Jan 23 at 7:34
• Note that $|\log(z+ia)|\sim \log(|z+ia|)\sim\log(|z|)$ and $|\frac{1}{1+z^2}|\sim |z|^{-2}$ (as $|z|\to\infty$), so we can bound $$\big|\int_{C_R}f(z)dz\big|\leq CR\frac{\log(R)}{R^2}\to0$$ – leoli1 Jan 23 at 9:11

$$I(a):=\int_{-\infty}^{\infty} \frac{\ln\sqrt{x^2+a^2}}{1+x^2}dx= \int_0^{\infty} \frac{\ln(x^2+a^2)}{1+x^2}dx$$ $$\frac{dI}{da}=\int_0^{\infty} \frac{2a}{(1+x^2)(x^2+a^2)}dx$$ $$\int \frac{2a}{(1+x^2)(x^2+a^2)}dx=\frac{2}{a^2-1}\tan^{-1}(\frac{x}{a})-\frac{2a}{a^2-1}\tan^{-1}(x)$$ $$\frac{dI}{da}=\frac{2}{a^2-1}\frac{\pi}{2}-\frac{2a}{a^2-1}\frac{\pi}{2}$$ After simplification : $$\frac{dI}{da}=\frac{\pi}{a+1}$$ $$I(a)=\pi\ln(a+1)+C$$ In order to determine $$C$$ we compute the integral for a particular value of $$a$$, for example $$a=0$$ : $$I(0)=\int_0^{\infty} \frac{\ln(x^2)}{1+x^2}dx$$ Change of variable $$\quad x=\frac{1}{t}$$ : $$I(0)=\int_{\infty}^0 \frac{\ln(\frac{1}{t^2})}{1+\frac{1}{t^2}}(-\frac{dt}{t^2})= -\int_0^{\infty} \frac{-\ln(t^2)}{t^2+1}(-dt)=-\int_0^{\infty} \frac{\ln(t^2)}{t^2+1}dt$$ This implies $$\int_0^{\infty} \frac{\ln(x^2)}{1+x^2}dx=-\int_0^{\infty} \frac{\ln(t^2)}{1+t^2}dt=0\quad\text{thus}\quad I(0)=0.$$ $$I(0)=\pi\ln(0+1)+C\quad\implies\quad C=0$$ $$\boxed{I(a):=\int_{-\infty}^{\infty} \frac{\ln\sqrt{x^2+a^2}}{1+x^2}dx=\pi\ln(a+1)}$$

I fact the integral can be evaluated by means of contour integrating in the complex plane. Let' consider the following closed contour C in the upper half-plane with the cut from $$x=ia$$ to $$x=0$$. We also choose the integrand $$f(x)=\frac{\log(x^2+a^2)}{1+x^2}$$ on the positive part of axis (right side). For the convenience of calculations we take $$a<1$$.

$$x=i$$ is a simple pole, and $$x=ia$$ is a logarithm branch point.

Due to the cut our integrand is a single-valued function in the upper half-plane. We integrate from $$x=0$$ along the positive axis, then along a big circle (radius R) counter clockwise, then along negative part of axis, from $$x=0$$ to $$x=ia$$, clockwise around the point $$x=ia$$ (circle of radius $$r$$) and then from $$x=ia$$ along the second bank of the cut to the initial starting point. It is easy to see that the integral along the big circle $$\to0$$ as soon as $$R\to{\infty}$$, and so does the integral around the point $$x=ia$$ when $$r\to0$$ . In the left quarter of half-plane (on the negative part of axis) the function $$(x^2+a^2)$$ will get the factor $$\exp(2\pi{i})$$ due to the logarithm branch point $$x=ia$$ (which we go around in the positive direction when integrating along a big circle).

Taking all together we get

$$\oint_C\frac{\log(x^2+a^2)}{1+x^2}$$=$$\int_0^{\infty}\frac{\log(x^2+a^2)}{1+x^2}dx+\int_R+\int_{-\infty}^0\frac{\log[(x^2+a^2)\exp(2\pi{i})]}{1+x^2}dx$$+$$\int_0^{ia}\frac{\log[(x^2+a^2)\exp(2\pi{i})]}{1+x^2}dx+\int_r +$$ $$+\int_{ia}^{0}\frac{\log(x^2+a^2)}{1+x^2}dx=2\pi{i}Res_{x=i}\frac{\log(x^2+a^2)}{1+x^2}$$

Setting $$R\to{\infty}$$ and $$r\to0$$, and taking into consideration that there are integrals cancelling each other, we get:

$$2I=\int_{-\infty}^{\infty}\frac{\log(x^2+a^2)}{1+x^2}dx=-\int_{-\infty}^{0}\frac{2\pi{i}}{1+x^2}dx-\int_0^{ia}\frac{2\pi{i}}{1+x^2}dx+2\pi{i}Res_{x=i}\frac{\log(x^2+a^2)}{1+x^2}$$.

When calculating the residual we have to bear in mind that $$\log(a-1)=\log[\exp(\pi{i}(1-a)]=\pi{i}\log(1-a)$$- due to the choice of logarithm branch (and the choice $$a<1$$).

Finally we get $$2I=-{\pi}^2i+2\pi\int_0^{a}\frac{1}{1-x^2}dx+\pi\log(a^2-1)=\pi\log(\frac{1+a}{1-a})+\pi\log[(1+a)(1-a)]$$.

$$I(a)=\int_{-\infty}^{\infty} \dfrac{\text{log}{\sqrt{x^2+a^2}}}{1+x^2}=\frac{1}{2}\int_{-\infty}^{\infty} \dfrac{\text{log}(x^2+a^2)}{1+x^2}=\pi\log(1+a); a<1$$

Calculations for $$a>1$$ can be done in the same fashion.