(Context) I tried out this proof I made up for fun:
Let $a \geq 0.$ Prove that $$\int_{-\infty}^{\infty}\frac{\exp\left(-\sin\left(ax\right)\right)\sin\left(\cos\left(ax\right)\right)}{x^{2}+1}dx = \pi\sin\left(\frac{1}{e^{a}}\right).$$
I put this integral and my answer in Desmos with a slider $a$, and it seems like Desmos is numerically approximating both expressions such that they're both off by a bit, probably around $0.1$ or $0.01$. (Also, I was treating $\infty$ as a huge number like $3000$ because Desmos sometimes has trouble dealing with it.)
(Attempt) Let that integral equal $I$. Rewriting it gives us
$$\int_{-\infty}^{\infty}\frac{\exp\left(-\sin\left(ax\right)\right)\sin\left(\cos\left(ax\right)\right)}{x^{2}+1}dx = \Im \int_{-\infty}^{\infty}\frac{\exp\left(ie^{iaz}\right)}{z^{2}+1}dz.$$
Let $f(z) = \frac{\exp\left(ie^{iaz}\right)}{z^{2}+1}$ and traverse counterclockwise along this curve $C:= \left[-R,R\right] \cup \Gamma$, where $C$ is a semicircle on and above the real axis in the complex plane and $R$ is a large radius approaching $\infty$. The set of singularities is $\left\{i,-i\right\}$ (after setting the denominator equal to $0$).
Notice that $i$ lives in $C$, so we can make it our pole. By Cauchy's Residue Theorem, we can rewrite $\oint_C f(z)dz$ as
$$2\pi i\text{Res}(f(z), z=i) = \int_{\Gamma}f(z)dz + \int_{-R}^R f(z)dz.$$
Solving the residue, we get
$$\text{Res}\left(\frac{\exp\left(ie^{iaz}\right)}{z^{2}+1}, z=i\right) = \frac{\exp{(ie^{ia(i)})}}{\frac{d}{dz}\left(z^2+1\right)\Big|_{z=i}} = \frac{\exp\left(ie^{-a}\right)}{2i}.$$
As for the contour integral, it converges to $0$ as $R \to \infty$. Notice if $z$ is on $\Gamma$, then $|z| = R$. Observe from using one of the Triangle Inequalities that
$$\left|\frac{\exp\left(ie^{iaz}\right)}{z^{2}+1}\right|\le\frac{\left|\exp\left(ie^{iaz}\right)\right|}{\left|\left|z\right|^{2}-1\right|}=\frac{1}{R^{2}-1}.$$
Using the ML-Inequality, observe that
$$0 \leq \left|\int_{\Gamma}f(z)dz\right| \leq |f(z)|\left(\frac{2\pi R}{2}\right) \leq \frac{\pi R}{R^2-1}.$$
Taking $R\to\infty$, we can use the Squeeze Theorem to get
$$\lim_{R\to\infty}\left|\int_{\Gamma}f(z)dz\right| = 0,$$
which implies
$$\lim_{R\to\infty}\int_{\Gamma}f(z)dz = 0.$$
Going back to $\oint_{C} f(z)dz$, we take $R\to\infty$ and $\Im$ on both sides to get
$$
\eqalign{
\Im\lim_{R\to\infty} 2\pi i\text{Res}(f(z), z=i) &= \Im\lim_{R\to\infty} \int_{\Gamma}f(z)dz + \Im\lim_{R\to\infty} \int_{-R}^R f(z)dz \cr
\Im\lim_{R\to\infty}\int_{-R}^{R}f(z)dz &= I \cr
&= \Im\left(\pi \exp{\left(ie^{-a}\right)}\right) \cr
&= \pi\sin{\left(\frac{1}{e^a}\right)}. \cr
}
$$
(Question) Is there something wrong with my proof? I've checked this over and over and I want to say there's nothing wrong and that I didn't miss anything trivial. Still, the approximations Desmos is giving me are sort of bothering me, even though I know Desmos isn't always reliable when evaluating things at $\infty$. And also, the proof doesn't seem to hold true if $a$ is negative. If my proof is correct, then how come it breaks when $a$ is negative?
If someone can shed some light, that would be great. Thank you in advance.
Best Answer
There is nothing wrong with your proof except that you have the incorrect upper bound on the magnitude of $ \exp \left(ie^{iaz} \right) $ in the upper half-plane.
While $\left| \exp \left(ie^{iaz} \right) \right|, \, a\ge 0, $ can exceed $1$ in the upper half-plane, it will never exceed $e$.
Let $z= x +iy$, where $x \in \mathbb{R}$ and $y >0$.
Then since $a \ge 0$ we have
$$ \begin{align} \left| \exp \left(ie^{iaz} \right) \right| &= \left| \exp \left(ie^{ia(x+iy)} \right) \right| \\ &=\left|\exp \left(i \left( \cos(ax) + i \sin(ax)\right)e^{-ay} \right) \right| \\ &= \left|\exp \left(i \cos(ax) e^{-ay} \right) \right| \exp \left( -\sin(ax) e^{-ay} \right)\\ &= \exp \left(-\sin(ax) e^{-ay} \right) \\ &\le \exp \left(e^{-ay} \right) \\ &\le \exp(1). \end{align}$$
On the other hand, if $a<0$, then there is no upper bound on the magnitude of $\exp \left(ie^{iaz} \right)$ in the upper half-plane.