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

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, \begin{equation} \displaystyle\int_{C_1} f(z) dz + \displaystyle\int_{C_R} f(z) dz = 2\pi i \text{Res}(f, i). \end{equation}

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.

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    $\begingroup$ 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 $|\log (R\cos \theta+i(R\sin \theta +a))|$? $\endgroup$ – ancientmathematician Jan 23 at 7:34
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    $\begingroup$ 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 $$\big|\int_{C_R}f(z)dz\big|\leq CR\frac{\log(R)}{R^2}\to0$$ $\endgroup$ – leoli1 Jan 23 at 9:11
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$$I(a):=\int_{-\infty}^{\infty} \frac{\ln\sqrt{x^2+a^2}}{1+x^2}dx= \int_0^{\infty} \frac{\ln(x^2+a^2)}{1+x^2}dx$$ $$\frac{dI}{da}=\int_0^{\infty} \frac{2a}{(1+x^2)(x^2+a^2)}dx$$ $$\int \frac{2a}{(1+x^2)(x^2+a^2)}dx=\frac{2}{a^2-1}\tan^{-1}(\frac{x}{a})-\frac{2a}{a^2-1}\tan^{-1}(x)$$ $$\frac{dI}{da}=\frac{2}{a^2-1}\frac{\pi}{2}-\frac{2a}{a^2-1}\frac{\pi}{2}$$ 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 $$\int_0^{\infty} \frac{\ln(x^2)}{1+x^2}dx=-\int_0^{\infty} \frac{\ln(t^2)}{1+t^2}dt=0\quad\text{thus}\quad I(0)=0.$$ $$I(0)=\pi\ln(0+1)+C\quad\implies\quad C=0$$ $$\boxed{I(a):=\int_{-\infty}^{\infty} \frac{\ln\sqrt{x^2+a^2}}{1+x^2}dx=\pi\ln(a+1)}$$

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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.

enter image description here

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.

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