Prove that this complex integral converges

Question

Let $a\in \mathbb C$ and $c\in \mathbb R$ such that $c\gt$Re($a$). Let $t\gt0,$ being a fixed constant.

Then show that the integral $$\lvert\int_{-\infty}^\infty\frac{e^{iyt}}{c+iy-a}dy\rvert\lt\infty,$$i.e. this integral conditionally converges.

My observation: Since $c\gt$Re($a$), the integrand is bounded around $y=0.$ And since $ \lvert e^{iyt}\rvert\le1$ and the norm of the denominator goes to $\infty$ as $y$ goes to $\pm \infty$, the integrand decays like $\frac{1}{y}$ at $\infty$. But in the real case, for example, $\int_{-\infty}^\infty\frac{1}{k+y}dy$ diverges, where $k$ is a fixed positive constant. So I guess there must be some differences between real integrals and complex integrals which I can't see. (Sorry I don't have a solid grounding in complex integrals...)

Thanks for help.

Answer 1

A direct real variables solution:

For $B>A >0$ Let $f(A,B)=\int_{A}^B\frac{e^{iyt}}{c+iy-a}dy$ It is enough to show that $|f(A,B)| \to 0, A,B \to \infty$ as then by symmetry we do the same for $C<D <0, C,D \to -\infty$ and use that Cauchy implies convergence theorem for integrals of say continuos functions.

Integrating by parts: $f(A,B)=\frac{1}{it}(\frac{e^{itB}}{c+iB-a}-\frac{e^{itA}}{c+iA-a})+\frac{1}{t}\int_{A}^B\frac{e^{iyt}}{(c+iy-a)^2}dy$ and it is then obvious that $|f(A,B)| =O(\frac{1}{B}+\frac{1}{A}+\int_{A}^B{\frac{1}{y^2}}dy)=O(\frac{1}{B}+\frac{1}{A})$ so we are done!

Answer 2

Let $f(z)=\frac{e^{zt}}{z-(a-c)}.$ Then, $f(z)$ is meromorphic with a simple pole at $a-c$ with residue $e^{(a-c)t}$, and we can apply the Residue theorem to get that

$$ \int_{\gamma} f(z)\textrm{d}z=2\pi i e^{(a-c)t} $$ for all Jordan curves $\gamma$ surrounding $a-c$. Let $\gamma_n$ be the curve which traces $i\mathbb{R}$ from $-in$ to $in$ and then traces the circle of radius $n$. For $n\geq |a-c|$, we see that

$$ \int_{-n}^n \frac{e^{iyt}}{c+iy-a}\textrm{d}y=2\pi ie^{(a-c)t}-\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} ne^{i\theta}\frac{e^{ne^{i\theta}}}{c+ne^{i\theta}-a}\textrm{d}\theta $$

Now, for $n\geq 2|a-c|$, we have

$$ \left|\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} ne^{i\theta}\frac{e^{ne^{i\theta}}}{c+ne^{i\theta}-a}\textrm{d}\theta\right|\leq \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \frac{n e^{n\cos(\theta)}}{n-|a-c|}\textrm{d}\theta\leq 2\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} e^{n\cos(\theta)}\textrm{d}\theta=4\int_0^{\frac{\pi}{2}} e^{-n\sin(\theta)}\textrm{d}\theta $$

Finishing the analysis, $$ \int_{\arcsin(\frac{1}{\sqrt{n}})}^{\frac{\pi}{2}}e^{-n\sin(\theta)}\textrm{d}\theta\leq \int_{\arcsin(\frac{1}{\sqrt{n}})}^{\frac{\pi}{2}}e^{-\sqrt{n}}\textrm{d}\theta\leq\frac{\pi}{2} e^{-\sqrt{n}}\to 0 $$ while $$ \int_0^{\arcsin(\frac{1}{\sqrt{n}})} e^{-n\sin(\theta)}\textrm{d}\theta\leq \arcsin(\frac{1}{\sqrt{n}})\to 0, $$ implying, all in all, that $\int_{-n}^n \frac{e^{iyt}}{c+iy-a}\textrm{d}y$ converges with limit $2\pi i e^{(a-c)t}$. Since $f(iy)\to 0$ for $|y|\to \infty$, we get that the integral converges.