# $\int_{-\infty}^\infty \frac{\sin(x-\frac 1x)}{x+\frac 1x}dx$ via complex analysis

I'm a little new to complex analysis, so bear with me:

First, I defined a function $$f(x)= \frac{\sin(x-\frac 1x)}{x+\frac 1x}$$, so that I could define a new function $$f(z)= \frac{e^{i(z-\frac 1z)}}{z+\frac 1z}$$, which I then simplified to be $$f(z)= \frac{ze^{i(z-\frac 1z)}}{z^2+1}$$

Second, I defined a contour $$C$$ that contains the interval $$[-R, -\epsilon]$$, the contour $$\gamma$$ which a a semicircle with radius $$\epsilon$$ centered at the origin, the interval $$[\epsilon, R]$$, and the semicircular contour $$\Gamma$$ with radius $$R$$ centered at the origin. I want to take the limit as $$R\rightarrow\infty$$ and $$\epsilon\rightarrow0$$

Thus, $$\int_C f(z)dz=\int_{-R}^{-\epsilon} f(z)dz+\int_\gamma f(z)dz+\int_\epsilon^R f(z)dz+\int_\Gamma f(z)dz$$

For the leftmost integral, I can simply use the residue theorem, as $$C$$ is a closed contour with a simple pole at $$i$$. Thus, $$\int_C f(z)dz=2\pi i Res(f, i)$$ $$=2\pi i\frac{1}{2e^2}=\frac{\pi i}{e^2}$$ Next, looking at the first and third integrals on the right, substitute $$z=-u$$ and $$dz=-du$$ into the first integral, so then $$\int_{-R}^{-\epsilon} f(z)dz+\int_\epsilon^R f(z)dz$$ $$\int_{\epsilon}^{R} \frac{-ue^{i(\frac 1u-u)}}{u^2+1}du+\int_\epsilon^R \frac{ze^{i(z-\frac 1z)}}{z^2+1}dz$$ which can then be combined to get $$\int_{\epsilon}^{R} \frac{z(e^{i(z-\frac 1z)}-e^{-i(z-\frac 1z)})}{z^2+1}dz$$ Next, multiplying by $$\frac{2i}{2i}$$, we get $$2i\int_{\epsilon}^{R} \frac{z\sin(z-\frac 1z)}{z^2+1}dz$$ And then, letting $$R\rightarrow\infty$$ and $$\epsilon\rightarrow0$$ we get $$2i\int_{0}^{\infty} \frac{z\sin(z-\frac 1z)}{z^2+1}dz$$ Also, since this integral is only on the real line, we can exchange the $$z$$ for an $$x$$, and also simplify it so that it looks like the original $$2i\int_{0}^{\infty} \frac{\sin(x-\frac 1x)}{x+\frac 1x}dx$$ which then leaves us with $$\frac{\pi i}{e^2}=2i\int_{0}^{\infty} \frac{\sin(x-\frac 1x)}{x+\frac 1x}dx+\int_\gamma f(z)dz+\int_\Gamma f(z)dz$$ The problem is that I'm not quite sure how to deal with the other two integrals in the equation. I pretty sure that the integral over $$\Gamma$$ tends toward 0 just by using the M-L inequality, but I'm not so sure about how to evaluate the integral over $$\gamma$$

• Why not using elementary analysis, e.g. Taylor Expansion of the sine and then getting a close expression for the indefinite integral of the terms? – fwgb Aug 17 at 20:25
• Reason being is that I’m doing this to try and better understand complex analysis and need the practice. Of course I could use a Taylor expansion, but I want to know how exactly I would do it this way. – alephnull14177 Aug 17 at 21:38

for the contour $$\Gamma$$ we parametrize: $$\eta : [0,1] \to \mathbb{C} : t \mapsto R e^{ti \pi}$$. Then $$\int_\Gamma f(z) ~dz= \int_0^1 \frac{\exp\big({i (R e^{ti \pi} - e^{-ti \pi}/R)}\big)}{Re^{ti \pi}+ e^{-ti \pi}/R}\cdot i \pi R e^{ti \pi}~dz$$Now for the enumerator we have $$\exp\big({i (R e^{ti \pi} - e^{-ti \pi}/R)}\big)=\\\exp\big( i R \cos (t \pi)\big) \cdot \exp\big(- R \sin (t \pi)\big)\cdot\exp\big(-i\cos(-t \pi)/R\big)\cdot \exp\big( \sin(-t\pi)/R\big).$$The second term hast is the important one. The ones with purely imaginary argument in the $$\exp$$ are of absolute value one. The last one tends to $$1$$ for $$R\to \infty$$ and as $$-R \sin(t\pi)$$ tends to $$- \infty$$ (because on $$[0,\pi]$$ we have $$\sin \ge 0$$) the second term goes to zero exponentially, overweighing the $$R$$ which we have from the derivative.
The denominator is bounded by $$2R$$ hence we get that the limit of the integral is zero. The contour $$\gamma$$ can be handled the same way, the trick again is writing $$\exp\big(e^{it}\big)=\exp\big( i \cos(t) - \sin(t) \big)$$.