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I'm trying to show $$PV \int_0^\infty\frac{\cos(\ln x)}{x^2+1}\,dx=\frac{\pi}{2\cosh(\pi/2)}.$$ My textbook says to do this by "integrating $e^{i\ln z}/(z^2-1)$ around a contour like Figure 7.3 but rotated 90◦ clock- wise so the straight side is along the y axis."

I took the original integral and reformulated it like this: $$I=\int_0^\infty\frac{\cos(\ln x)}{x^2+1}\,dx=\frac{1}{2}\text{ Re}\left[\int_{-\infty}^\infty\frac{e^{i\ln x}}{x^2+1}\,dx\right]$$

Next I set up the contour integral $$\oint_C\frac{e^{i\ln z}}{(z^2-1)}\,dz=2\pi i\sum Res$$

With $\sum Res = Res[z=1]$.

I then managed to show that the semicircular paths around the contour go to zero as $R$ and $\epsilon$ go to infinity and zero respectively. This implies the contour integral is equal to the principal value of $f(z)$ evaluated from $-\infty$ to $\infty$.

But my problem is that $\frac{1}{2}$ of the real part of the sum of the residues does not equal $\frac{\pi}{2\cosh(\pi/2)}$.

Any pointers would be very much appreciated, I've been working on this for quite a while now!

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    i don't know where exactly you have gone wrong but want to ask you two questions : the residues are at $z=\pm i$ right? Furthermore how you dealt with the branch cut of $\log(z)$? – 2017-02-09
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    if i where you, i would substitute $x=e^q$. the resulting integral can be obtained by integrating a rectangular contour of width $i \pi$ – 2017-02-09
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    @tired I would have assumed to use those residues but for some reason the book wants me to integrate the function with residues at +- 1. Perhaps it's a mistake? – 2017-02-09

3 Answers 3

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According to your text book we consider $f(z)=\dfrac{e^{i\log z}}{z^2+1}.$
Integrating $f(z)$ around a contour in Figure 7.3, we have \begin{align} \left(\int_{-R}^{-\varepsilon }+\int_{C_\varepsilon }+\int_{\varepsilon }^R +\int_{C_R}\right)f(z)dz=2\pi i\operatorname{Res}(f; i), \end{align} since $f(z)$ has a simple pole at $z=i$. It is easy to see that $$ \operatorname{Res}(f; i)=\frac{ e^{-\frac{\pi}{2}} } {2i} . $$ Since $$ \int_{-R}^{-\varepsilon }\dfrac{e^{i\log z}}{z^2+1}dz=-\int_{R}^{\varepsilon }\dfrac{e^{i(\log x+\pi i)}}{x^2+1}dx=e^{-\pi}\int_{\varepsilon }^R\dfrac{e^{i\log x}}{x^2+1}dx, $$

we have $$ \left(\int_{-R}^{-\varepsilon }+\int_{\varepsilon }^R\right)f(z)dz=\left(1+e^{-\pi}\right)\int_{\varepsilon }^R\dfrac{e^{i\log x}}{x^2+1}dx.$$ Therefore if we could show that$$ \lim_{\varepsilon \to 0} \int_{C_\varepsilon } f(z)dz=0\quad\text{and}\quad\lim_{R\to \infty}\int_{C_R}f(z)dz=0, $$ we have $$ \left(1+e^{-\pi}\right)\int_0^\infty\dfrac{e^{i\log x}}{x^2+1}dx=\pi e^{-\frac{\pi}{2}}.$$ Taking real parts of both sides we have $$ \int_0^\infty\dfrac{\cos(\log x)}{x^2+1}dx= \frac{\pi}{e^{\frac{\pi}{2}}+e^{-\frac{\pi}{2}}}.$$

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HINT :

The poles of $\frac{\cos(\ln z)}{z^2+1}$ are $z=i$ and $z=-i$ $$\text{Res}_{z=i}=-\frac{i}{2}\cosh\left(\frac{\pi}{2}\right)$$ $$\text{Res}_{z=-i}=\frac{i}{2}\cosh\left(\frac{\pi}{2}\right)$$ I suppose that you can take it from here.

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An alternative solution. We do not really need the $\text{PV}$ before the integral, since $\cos$ is a bounded function, hence $\frac{\cos\log x}{x^2+1}=\text{Re}\frac{x^i}{x^2+1}$ is integrable over $\mathbb{R}^+$. By setting $x=e^t$ we get: $$ I=\int_{0}^{+\infty}\frac{\cos\log x}{x^2+1}\,dx = \int_{-\infty}^{+\infty}\frac{\cos t}{2\cosh(t)}\,dt$$ and the last integral is easily solvable through the Fourier transform, leading to:

$$ I = \frac{\pi}{2}\int_{-\infty}^{+\infty}\frac{\delta(s-1)+\delta(s+1)}{2\cosh\left(\frac{\pi s}{2}\right)}\,ds=\color{red}{\frac{\pi}{2\cosh\frac{\pi}{2}}}$$ as wanted.