# Calculate the value of $\int_0^\infty \frac{\sqrt{x}\cos(\ln(x))}{x^2+1}\,dx$

I'm asked to evaluate the integral $$\displaystyle\int_0^\infty \frac{\sqrt{x}\cos(\ln(x))}{x^2+1}\,dx$$.

I tried defining a funcion $$f(z)=\frac{e^{(1/2+i)\operatorname{Log}(z)}}{z^2+1}$$, taking $$\operatorname{Log}$$ with a branch cut along the positive real axis: ($$\operatorname{Log}(z)=\ln(|z|)+i\arg(z))$$.

Using residue theorem with the "pacman" contour.

However when trying to bound the integral around a small circle around $$0$$, I cannot conclude it converges to $$0$$.

My attempt was $$|\int_{\gamma_\epsilon}f|\leq 2\pi\epsilon|e^{(0.5+i)(\ln|\epsilon|+i\theta))}|\frac{1}{\epsilon^2-1}\leq C\epsilon^{-0.5}.$$

I'd love it if someone could either suggest a different way to bound the integral around $$0$$ of this function, or maybe suggest an easier complex function to work with.

Edit:

The wonderful "Related" algorithm of this site managed to link me to this answer Looking at it , a more general statement is proved, but the proof fails when we have $$\alpha=0.5+i$$ (The circle around $$0$$ doesnt converge to $$0$$ by the proof given there, as a matter of fact any $$\alpha$$ with $$Re(\alpha)>0$$ would fail.)

• Are you forced to use complex analysis? – Frank W. Jan 25 at 18:24
• It's a practice problem for the residue theorem but Id like seeing different approaches aswell! – Sar Jan 25 at 18:27
• I could be wrong, but I doubt you want the branch cut directly on the path of integration! Usually, in complex analysis, you choose branch cuts to avoid the contour altogether. – Adrian Keister Jan 25 at 18:29
• You`re correct, using a contour of upper half circle with a branch cut along the negative imaginary axis would be more simple, but both yield the same result eventually (And both have the same problem of convergence around 0 ) – Sar Jan 25 at 19:07
• I tried to evaluate your integral using a different approach without using contour integration (see here) but got stuck on an infinite sum. – omegadot Jan 26 at 10:20

As @Adrian suggested, define $$\log z =\log |z|+i\arg(z)$$ where $$\arg(z)\in (0,2\pi)$$ and let the contour be a keyhole contour.

Then $$\left|\int_{\gamma_R}\frac{e^{(1/2+i)\log z}}{z^2+1}\,dz\right|\le \int_{\gamma_R}\frac{e^{1/2 \log|z|-\arg(z)}}{R^2-1}\,|dz|\le C\frac{R^{3/2}}{R^2-1}\stackrel{R\to\infty}\longrightarrow 0,$$ $$\left|\int_{\gamma_r}\frac{e^{(1/2+i)\log z}}{z^2+1}\,dz\right|\le \int_{\gamma_r}\frac{e^{1/2 \log|z|-\arg(z)}}{1-r^2}\,|dz|\le Cr^{3/2}\stackrel{r\to 0}\longrightarrow 0.$$ Thus it follows by residue theorem $$\lim_{\epsilon\to 0}\left(\int_{\gamma_\epsilon} f(z)dz +\int_{\gamma_{-\epsilon}} f(z)dz\right) =2\pi i\left(\text{res}_{z=i}f(z)+\text{res}_{z=-i}f(z)\right).$$ We find$$\lim_{\epsilon\to 0}\int_{\gamma_\epsilon} f(z)dz=\int_0^\infty \frac{\sqrt{x}e^{i\ln x}}{x^2+1}\,dx,$$ $$\lim_{\epsilon\to 0}\int_{\gamma_{-\epsilon}} f(z)dz=-\int_0^\infty \frac{e^{(1/2+i)(\ln x+2\pi i)}}{x^2+1}\,dx=+e^{-2\pi}\int_0^\infty \frac{\sqrt{x}e^{i\ln x}}{x^2+1}\,dx.$$ And also $$\text{res}_{z=i}f(z)=\frac{e^{(1/2+i)\frac{\pi i}{2}}}{2i}=\frac{e^{-\pi/2+\pi i/4}}{2i},$$ $$\text{res}_{z=-i}f(z)=-\frac{e^{(1/2+i)\frac{3\pi i}{2}}}{2i}=-\frac{e^{-3\pi/2+3\pi i/4}}{2i}.$$ Thus the given integral is $$\frac{\pi}{1+e^{-2\pi}}\Re\left(e^{-\pi/2+\pi i/4}-e^{-3\pi/2+3\pi i/4}\right)=\frac{\pi\cosh(\frac{\pi}{2})}{\sqrt{2}\cosh(\pi)}\sim 0.4805.$$ (I found that this value coincides with the integral numerically by wolframalpha.)

• This is exactly what I meant to do. Can you please elaborate how you bounded $\gamma_r$ ?( For the small $r$) In addition - I think the factor outside the integral of $\gamma_{-\epsilon}$ is supposed to be $e^{2\pi i(\frac{1}{2}+i)}$ – Sar Jan 25 at 19:29
• @Sar Since $r\to 0$, the term $1/(1-r^2)$ is bounded as $r\to 0$. And similarly, since $\arg(z)$ is bounded, $e^{\log|z|/2 -\arg(z)}$ is bounded by $Ce^{\log r/2}=C\sqrt{r}$. Finally, length of $\gamma_r$ is about $2\pi r$. This gives the bound $Cr^{3/2}$. I hope this makes it clear :) – Song Jan 25 at 19:32
• Yes, Much simpler than I thought . Thank you very much ! Is it true what I commented on the factor on the previous comment? Or am I missing something here? – Sar Jan 25 at 19:35
• @Sar Oh, you are also right. Since $e^{2\pi i(1/2+i)}=e^{\pi i-2\pi}=-e^{-2\pi}$, both results are correct. – Song Jan 25 at 19:37

Integrating by parts twice, we get $$\int_0^\infty\cos(x)\,e^{-ax}\,\mathrm{d}x=\frac{a}{a^2+1}\tag1$$ Therefore, \begin{align} \int_0^\infty\frac{\sqrt{x}\cos(\log(x))}{x^2+1}\,\mathrm{d}x &=\int_{-\infty}^\infty\frac{\cos(x)}{e^{2x}+1}e^{3x/2}\,\mathrm{d}x\tag2\\ &=\int_{-\infty}^\infty\frac{\cos(x)}{e^{2x}+1}e^{x/2}\,\mathrm{d}x\tag3\\ &=\int_0^\infty\frac{\cos(x)}{e^x+e^{-x}}\left(e^{x/2}+e^{-x/2}\right)\mathrm{d}x\tag4\\ &=\int_0^\infty\cos(x)\sum_{k=0}^\infty(-1)^k\left(e^{-(4k+1)x/2}+e^{-(4k+3)x/2}\right)\mathrm{d}x\tag5\\ &=\frac12\sum_{k=0}^\infty(-1)^k\left[\frac{k+\frac14}{\left(k+\frac14\right)^2+\frac14}+\frac{k+\frac34}{\left(k+\frac34\right)^2+\frac14}\right]\tag6\\[6pt] &=\frac\pi{\sqrt2}\frac{\cosh(\pi/2)}{\cosh(\pi)}\tag7 \end{align} Explanation:
$$(2)$$: substitute $$x\mapsto e^x$$
$$(3)$$: substitute $$x\mapsto-x$$
$$(4)$$: average $$(2)$$ and $$(3)$$ and apply symmetry
$$(5)$$: expand into power series
$$(6)$$: apply $$(1)$$
$$(7)$$: use $$(7)$$ from this answer