Evaluating the complex integral $\int_{0}^{\infty} \frac {\sin (\ln x) dx }{x^2 + 4} $

Question

I want to evaluate following integral:

$\int_{0}^{\infty} \frac {\sin (\ln x) dx }{x^2 + 4} $

• Obviously $x$ $ \gt $ $0$ and the function we want to integrate isn't even nor odd. And I need to avoid $0$.

• We got two first order poles at $+2i$ and $-2i$

• $f(z) = \frac {\sin (\ln z) }{z^2 + 4} = \Im \{\frac {e^{(i\ln z)} }{z^2 + 4} \} $ and we are dealing with a complex logarithm

• The residue of the function above for $+2i$ is equal to $res(f, +2i) =\frac {e^{(i Ln 2)}}{4i}$ and I am considering calculating the residue for $-2i$ (it would just switch a sign)

The solution is according to textbook: $\frac {\pi \sin (Ln(2)}{4 \cosh(\pi /2)}$

My questions:

1) How to deal with the integral from $0$ to $\infty$, not from $- \infty$ to $\infty$ in this case. I was used to deal with even functions where it was obvious.

2) I have no clue about the integration path (a semi-circle with branch cuts?)

3) Why that $cosh$ in that solution?

This type of the complex integral is very new to me and I would appreciate any help!

Answer 1

Use the substitution $$x = 2e^t$$ for instance. This gives: $$ \begin{aligned} \int_{0}^{\infty} \frac {\sin \ln x }{x^2 + 4}\; dx &= \int_{-\infty}^{\infty} \frac {\sin (\ln 2 + t) }{4e^{2t} + 4}\; 2e^t\; dt \\ &= \frac 12 \int_{-\infty}^{\infty} \frac {\sin \ln 2\cos t+\cos \ln 2\sin t }{e^{t} + e^{-t}}\; dt \\ &= \frac 12 \int_{-\infty}^{\infty} \frac {\sin \ln 2\cos t +\text{odd function}}{e^{t} + e^{-t}}\; dt \\ &= \frac 12\sin\ln 2 \int_{-\infty}^{\infty} \frac {\cos t}{e^{t} + e^{-t}}\; dt \ . \end{aligned} $$ This is a better situation suited for an application of the residue theorem. The residues of the function $$ f(z)=\frac{\cos z}{e^z + e^{-z}} $$ in the poles $$ k\cdot \frac {i\pi} 2\ ,\qquad\text{ $k$ odd, }k=2n+1\ , $$ are correspondingly $$ -\frac i2(-1)^n\cosh \frac {k\pi}2\ . $$ Now we need to find the "good contour". The known result already tells us that $i\pi/2$ counts, but the "next" residue, $3i\pi/2$ "should not count". OK, let us then consider the rectangular contour with the corners $$ -R\ , \ +R\ , \ +R+i\pi\ , \ -R+i\pi\ . $$ Then for $z\in[-R,R]$ $$ \Re f(z+i\pi) = \Re \frac{\cos (z+i\pi)}{e^{z+i\pi} + e^{-z-i\pi}} = \Re \frac{\cos z\cos(i\pi)-\sin z\sin(i\pi)}{-(e^{z} + e^{-z})} = \Re \frac{\cos z\cosh \pi}{-(e^{z} + e^{-z})} =- \cosh \pi f(z)\ . $$ Applying the residue theorem on the rectangle: $$ \begin{aligned} 2\pi\,i\cdot \left(-\frac{1}{2} i \, \cosh\left(\frac\pi{2}\right)\right) &= \lim_{R\to \infty}\int_{-R}^R f(z)\; dz + \lim_{R\to \infty}\int_{R+i\pi}^{-R+i\pi} f(z)\; dz \\ &= (1+\cosh\pi)\lim_{R\to \infty}\int_{-R}^R f(z)\; dz \\ &= 2\cosh^2\left(\frac \pi2\right)\cdot\int_{-\infty}^\infty f(z)\; dz\ . \end{aligned} $$ It remains to put all together in one line: $$ \int_{0}^{\infty} \frac {\sin \ln x }{x^2 + 4}\; dx = \frac 12\sin\ln2\int_{\Bbb R}f = \frac 12\sin\ln2\cdot \frac \pi{2\cosh(\pi/2)}\ . $$ $\square$

Answer 2

Consider the function $f(z) = \frac{e^{i\log z}}{4+z^2}$ where the branch of the logarithm corresponds to $-\pi < \arg z \leq \pi$. We will integrate $f(z)$ around the following "key-hole" contour:

$C_R$ is a circle of radius $R$ and $C_{\epsilon}$ is a half-circle of radius $\epsilon$. Both of them are centered at $0$.

As $R\to\infty$ and $\epsilon\to 0^+$, the integrals around $C_R$ and $C_\epsilon$ tend to $0$. So, we are only left with the integrals above and below the branch cut.

While calculating the residues, one must be careful about the branch of the logarithm.

\begin{align*} \mathop{\text{Res}}\limits_{z=2i} \; f(z)&= \lim_{z\to 2i} (z-2i) \frac{e^{i\log z}}{z^2+4} \\ &= \frac{e^{i\log(2i)}}{4i} \\ &= \frac{e^{i\ln 2- \arg (i)}}{4i}\\ &= \frac{e^{i\ln 2 -\frac{\pi}{2}}}{4i} \end{align*}

Similarly, \begin{align*} \mathop{\text{Res}}\limits_{z=-2i} \; f(z)=-\frac{e^{i\ln 2 -\arg(-i)}}{4i} =-\frac{e^{i\ln 2 +\frac{\pi}{2}}}{4i} \end{align*}

Therefore, using the Residue Theorem,

\begin{align*} \int_{-\infty}^0 \frac{e^{i(\ln|x|+i\pi)}}{4+x^2}dx +\int_0^{-\infty} \frac{e^{i(\ln|x|-i\pi)}}{4+x^2}dx &= 2\pi i \left(\mathop{\text{Res}}\limits_{z=2i} \; f(z) + \mathop{\text{Res}}\limits_{z=-2i} \; f(z)\right) \\ \Rightarrow e^{-\pi}\int_0^\infty \frac{e^{i\ln x}}{4+x^2}dx - e^{\pi}\int_0^\infty \frac{e^{i\ln x}}{4+x^2}dx &= 2\pi i \left( \frac{e^{i\ln 2}}{4i}e^{-\frac{\pi}{2}}- \frac{e^{i\ln 2}}{4i}e^{\frac{\pi}{2}}\right) \\ \Rightarrow -2\sinh(\pi) \int_0^\infty \frac{e^{i\ln x}}{4+x^2}dx &= -\sinh\left(\frac{\pi}{2}\right) \pi e^{i\ln 2} \\ \Rightarrow \int_0^\infty \frac{e^{i\ln x}}{4+x^2}dx &= \frac{\pi e^{i\ln 2}}{4\cosh\left(\frac{\pi}{2}\right)} \end{align*} Now, separate the imaginary parts to get the answer.

Answer 3

$$ \begin{align} \int_0^\infty\frac{\sin(\log(x))\,\mathrm{d}x}{x^2+4} &=\frac14\int_{-\infty}^\infty\frac{\sin(x+\log(2))\,\mathrm{d}x}{\cosh(x)}\tag1\\ &=\frac{\sin(\log(2)}4\int_{-\infty}^\infty\frac{\cos(x)\,\mathrm{d}x}{\cosh(x)}\tag2\\[3pt] &=\frac\pi4\sin(\log(2))\,\mathrm{sech}\!\left(\frac\pi2\right)\tag3 \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto2e^x$
$(2)$: the even part of $\sin(x+\log(2))$ is $\sin(\log(2))\cos(x)$
$(3)$: double the result of this answer

Answer 4

An alternative approach. By Euler's Beta function and the reflection formula for the $\Gamma$ function we have $$ \int_{0}^{+\infty}\frac{x^a}{x^2+4}\,dx = \frac{\pi}{2^{2-\alpha}\cos\frac{\pi\alpha}{2}}\tag{1}$$ for any $\alpha$ such that $\text{Re}(\alpha)\in(-1,1)$. By considering $\alpha=i$ we get $$ \int_{0}^{+\infty}\frac{\cos\log(x)+i\sin\log(x)}{x^2+4}\,dx=\frac{\pi}{4\cosh\frac{\pi}{2}}\left[\cos\log 2+i\sin\log 2\right]\tag{2} $$ so $$ \int_{0}^{+\infty}\frac{\sin\log(x)}{x^2+4}\,dx=\frac{\pi\sin\log 2}{4\cosh\frac{\pi}{2}}.\tag{3} $$