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I need to calculate the integral of $$\int_{0}^{\infty} \,\frac{\sqrt{x}\,\cos\big(\ln(x)\big)}{x^2+1}\,\text{d}x \,.$$ I tried to use the complex function of $$f(z) = \frac{\sqrt{z}\cos\big(\text{Log}(z)\big)}{z^2+1} $$ and the contour of Pacman, or half a disk in the upper half plane (of course, taking into consideration the residues and using the Residue Theorem), without any success. Any help would be appreciated.

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    $\begingroup$ You may find a general expression for $\int_{0}^{+\infty}\frac{x^\alpha}{x^2+1}\,dx$ (under suitable constraints for $\alpha$) then consider what you get by considering the real part of the previous integral for $\alpha=\frac{1}{2}+i$. $\endgroup$ – Jack D'Aurizio Jul 29 '18 at 21:09
  • $\begingroup$ Wow, sounds interesting. But I didn't understand how I connect the element of $ x^i $ and $ cos(ln(x)) $ :\ $\endgroup$ – user500982 Jul 29 '18 at 21:30
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    $\begingroup$ $$\cos(\log x)=\text{Re}\,e^{i\log x} = \text{Re}(x^i).$$ $\endgroup$ – Jack D'Aurizio Jul 30 '18 at 1:16
  • $\begingroup$ To close voters, the OP did indicate that he was trying something. $\endgroup$ – Batominovski Jul 30 '18 at 8:10
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A general hint: never try to use trigonometric functions in this kind of exercises. In complex numbers it is very hard to find bounds for sine and cosine on a disk or on half a disk. So in the complex function you defined you need to replace cosine with exponential, and in the end just take the real or imaginary part, using the fact that $e^{iz}=cos(z)+isin(z)$.

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  • $\begingroup$ Thanks for the answer! I actually tried to use the function $ f(z) = \frac{\sqrt{z}\exp(i*Log(z))}{z^2+1} $ , but when I took the curve of half a disk (with a "bump" at 0), I had troubles bounding it on $ |z| = R $ where $ R $ goes to $ \infty $ because of the exponent. How can I bound it on both half disks ($ |z| = R $ and $ |z| = \epsilon $)? $\endgroup$ – user500982 Jul 29 '18 at 21:03
  • $\begingroup$ Exponentials are bounded in the complex plane $\endgroup$ – Davide Morgante Jul 29 '18 at 21:08
  • $\begingroup$ @DavideMorgante I don't think this is true.. By liouville theorem they would have been constant, for example... It is the exponent of the real part... $\endgroup$ – user500982 Jul 29 '18 at 21:17
  • $\begingroup$ Oh yes, I'm sorry, for a moment I just thought about the imaginary part! $\endgroup$ – Davide Morgante Jul 29 '18 at 21:19
  • $\begingroup$ For this exercise I actually recommend to use the Pacman contour. Then you can bound the function on the $R$ disk by the maximum of $2πR|\frac{\sqrt{z}e^{ilog(z)}}{z^2+1}|$ on the disk. Now, on that disk you have $log(z)=R+i∗arg(z)$ and hence $e^{ilogz}=e^{iR}∗e^{−arg(z)}$.The absolute value of $e^{iR}$ is 1 (because $R$ is real) and arg(z) doesn't depend on $R$ at all. Hence your bound is the maximum of $2πR|\frac{\sqrt{z}}{z^2+1}|$ which clearly goes to zero when $R$ goes to infinity because the power of $R$ in the denominator is higher. $\endgroup$ – Mark Jul 29 '18 at 21:24
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Claim. Let $\alpha\in\mathbb{C}$ be such that $-1<\text{Re}(\alpha)<+1$. Then, $$\int_0^\infty\,\frac{x^\alpha}{x^2+1}\,\text{d}x=\frac{\pi}{2}\,\text{sech}\left(\frac{\text{i}\pi\alpha}{2}\right)=\frac{\pi}{2}\,\sec\left(\frac{\pi\alpha}{2}\right)\,.$$ Here, $x^\alpha$ is defined to be $\exp\big(\alpha\,\ln(x)\big)$ for all $x>0$.

Without loss of generality, we assume that $\text{Im}(\alpha)\geq 0$. That is, $\alpha=\beta+\text{i}\,\gamma$, where $\beta,\gamma\in\mathbb{R}$ with $-1<\beta<+1$ and $\gamma\geq0$. Use the negative imaginary line to be the branch cut of $z^\alpha$ in the definition of the function $f$ below: $$f(z):=\frac{z^{\alpha}}{z^2+1}\text{ for }z\in\mathbb{C}\setminus\big(\text{i}\,\mathbb{R}_{\leq 0}\cup\{-\text{i},+\text{i}\}\big)\,.$$

Pick a real number $\epsilon$ such that $0<\epsilon<1$, and choose the counterclockwise contour $$C_\epsilon=\left[\epsilon,\frac1\epsilon\right]\cup \Big\{\frac1\epsilon\,\exp(\text{i}\theta)\,\Big|\,\theta\in[0,\pi]\,\Big\}\cup\left[-\frac1\epsilon,-\epsilon\right]\cup\big\{\epsilon\,\exp(\text{i}\theta)\,\big|\,\theta\in[\pi,0]\big\}\,.$$ For a positive real number $r$, let $\Gamma_r$ denote the semicircular arc $\Big\{r\,\exp(\text{i}\theta)\,\Big|\,\theta\in[0,\pi]\,\Big\}$ (oriented in the counterclockwise direction). Thus, for $z=\frac{1}{\epsilon}\,\exp(\text{i}\theta)$, where $0\leq \theta\leq \pi$, we have $$\big|f(z)\big|\leq\frac{\Big|\exp\big(-\gamma\theta+\text{i}\,\beta\theta-\text{i}\,\gamma\,\ln(\epsilon)\big)\Big|}{\epsilon^\beta\,\left(\frac{1}{\epsilon^2}-1\right)}=\frac{\exp(-\gamma\theta)\,\epsilon^{2-\beta}}{1-\epsilon^2}\leq \frac{\epsilon^{2-\beta}}{1-\epsilon^2}\,.$$ As a consequence, $$\left|\int_{\Gamma_{\frac1\epsilon}}\,f(z)\,\text{d}z\right|\leq \frac{\pi\,\epsilon^{1-\beta}}{1-\epsilon^2}\underset{\small\epsilon\to0^+}{\longrightarrow}0\,.$$ Similarly, $$\left|\int_{\Gamma_{\epsilon}}\,f(z)\,\text{d}z\right|\leq \frac{\pi\,\epsilon^{1+\beta}}{1-\epsilon^2}\underset{\small\epsilon\to0^+}{\longrightarrow}0\,.$$

Ergo, $$\lim_{\epsilon\to0^+}\,\oint_{C_\epsilon}\,f(z)\,\text{d}z=\int_0^\infty\,\frac{x^\alpha}{x^2+1}\,\text{d}x+\int^{\infty}_0\,\frac{\exp(\text{i}\pi\alpha)\,x^\alpha}{x^2+1}\,\text{d}x\,.$$ Since $$\lim_{\epsilon\to0^+}\,\oint_{C_\epsilon}\,f(z)\,\text{d}z=2\pi\text{i}\,\text{Res}_{z=\text{i}}\left(\frac{z^\alpha}{z^2+1}\right)=2\pi\text{i}\,\left(\frac{\text{i}^\alpha}{2\text{i}}\right)=\pi\,\exp\left(\frac{\text{i}\pi\alpha}{2}\right)\,.$$ This shows that $$\int_0^\infty\,\frac{x^\alpha}{x^2+1}\,\text{d}x=\frac{\pi}{2}\,\text{sech}\left(\frac{\text{i}\pi\alpha}{2}\right)=\small\frac{\pi}{2}\,\left(\frac{\cos\left(\frac{\pi\beta}{2}\right)\,\cosh\left(\frac{\pi\gamma}{2}\right)+\text{i}\,\sin\left(\frac{\pi\beta}{2}\right)\,\sinh\left(\frac{\pi\gamma}{2}\right)}{\cos^2\left(\frac{\pi\beta}{2}\right)\,\cosh^2\left(\frac{\pi\gamma}{2}\right)+\sin^2\left(\frac{\pi\beta}{2}\right)\,\sinh^2\left(\frac{\pi\gamma}{2}\right)}\right)\,.$$

Corollary 1. Let $\beta$ and $\gamma$ be real numbers such that $|\beta|<1$. Then, $$\int_{0}^{\infty}\,\frac{x^\beta\,\cos\big(\gamma\,\ln(x)\big)}{x^2+1}\,\text{d}x=\pi\,\left(\frac{\cos\left(\frac{\pi\beta}{2}\right)\,\cosh\left(\frac{\pi\gamma}{2}\right)}{\cos(\pi\beta)+\cosh(\pi\gamma)}\right)\,.$$


Corollary 2. Let $\beta$ and $\gamma$ be real numbers such that $|\beta|<1$. Then, $$\int_{0}^{\infty}\,\frac{x^\beta\,\sin\big(\gamma\,\ln(x)\big)}{x^2+1}\,\text{d}x=\pi\,\left(\frac{\sin\left(\frac{\pi\beta}{2}\right)\,\sinh\left(\frac{\pi\gamma}{2}\right)}{\cos(\pi\beta)+\cosh(\pi\gamma)}\right)\,.$$


Corollary 3. Let $\alpha$ be a complex number such that $\big|\text{Re}(\alpha)\big|<1$. Then, $$\int_{-\infty}^{+\infty}\,\frac{x^\alpha}{x^2+1}\,\text{d}x=\pi\,\exp\left(\frac{\text{i}\pi\alpha}{2}\right)\,,$$ where $x^\alpha$ is defined to be $\exp\Big(\alpha\,\ln\big(x\big)\Big)$ when $x>0$, and $\exp\Big(\alpha\,\ln\big(|x|\big)+\text{i}\pi\alpha\Big)$ when $x<0$.


Corollary 4. Let $\beta$ and $\gamma$ be real numbers such that $|\beta|+|\gamma|<1$. Then, $$\int_{0}^{\infty}\,\frac{x^\beta\,\cosh\big(\gamma\,\ln(x)\big)}{x^2+1}\,\text{d}x=\pi\,\left(\frac{\cos\left(\frac{\pi\beta}{2}\right)\,\cos\left(\frac{\pi\gamma}{2}\right)}{\cos(\pi\beta)+\cos(\pi\gamma)}\right)\,.$$


Corollary 5. Let $\beta$ and $\gamma$ be real numbers such that $|\beta|+|\gamma|<1$. Then, $$\int_{0}^{\infty}\,\frac{x^\beta\,\sinh\big(\gamma\,\ln(x)\big)}{x^2+1}\,\text{d}x=\pi\,\left(\frac{\sin\left(\frac{\pi\beta}{2}\right)\,\sin\left(\frac{\pi\gamma}{2}\right)}{\cos(\pi\beta)+\cos(\pi\gamma)}\right)\,.$$


To answer the OP's question, write $$I:=\int_0^\infty\,\frac{\sqrt{x}\,\cos\big(\ln(x)\big)}{x^2+1}\,\text{d}x=\text{Re}\left(\int_0^\infty\,\frac{x^{\frac{1}{2}}\,\exp\big(\text{i}\,\ln(x)\big)}{x^2+1}\,\text{d}x\right)\,.$$ Thus, it suffices to find $$J:=\int_0^\infty\,\frac{x^{\frac{1}{2}}\,\exp\big(\text{i}\,\ln(x)\big)}{x^2+1}\,\text{d}x=\int_0^\infty\,\frac{x^{\frac{1}{2}}\,x^\text{i}}{x^2+1}\,\text{d}x=\int_0^\infty\,\frac{x^{\frac{1}{2}+\text{i}}}{x^2+1}\,\text{d}x\,.$$ From the work above, $$J=\frac{\pi}{2}\,\text{sech}\Biggl(\frac{\text{i}\pi}{2}\left(\frac{1}{2}+\text{i}\right)\Biggr)=\frac{\pi}{\sqrt{2}}\,\left(\frac{\cosh\left(\frac{\pi}{2}\right)+\text{i}\,\sinh\left(\frac{\pi}{2}\right)}{\cosh(\pi)}\right)\,.$$ This gives $$I=\text{Re}(J)=\frac{\pi\,\cosh\left(\frac{\pi}{2}\right)}{\sqrt{2}\,\cosh(\pi)}\approx0.48085\,.$$

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Hint

$$\cos(x)\rightarrow \Re(e^{iz})$$

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