Integral $J(\alpha):=\frac{1}{2i\pi}\int_\gamma \frac{\exp(\alpha\cdot z)}{z^2}dz$

Question

Let $a>0$ and $\gamma:t\to a+it,t\in\Bbb{R}$. Denote $$J(\alpha):=\frac{1}{2i\pi}\int_\gamma \frac{\exp(\alpha\cdot z)}{z^2}dz\quad\alpha\in\Bbb{R}$$

The exercise it to show that $J(\alpha)=0$ if $\alpha\le0$ and $J(\alpha)=\alpha$ if $\alpha\ge0.$

So first I want to show the integral converges, but if I am not mistaken I can write $$J(\alpha)=\frac1{2i\pi}\int_{-\infty}^\infty\frac{e^{\alpha(a+it)}}{(a+it)^2}dt$$ and using asymptotic behavior of $\frac{1}{(a+it)^2}$ I get that the limit as $t\to+\infty$ or $-\infty$ is zero.

Now to compute the integral I am not sure how can I do, mechanically I I consider $D_R:=[a-iR+a+iR]+\Gamma_R$ where $\Gamma_R$ is a half-circle of diameter $[a-iR+a+iR]$.

But now to compute the integral over $\Gamma_R$ I parametrize the circle as $t\to a+(a+iR)e^{it}$ so that I get $$\int_{\Gamma_R}\frac{\exp(\alpha\cdot z)}{z^2}dz=\int_{-\pi}^\pi \frac{exp\bigl(\alpha\cdot(a+(a+iR)e^{it})\bigr)}{(a+(a+iR)e^{it})^2}\cdot\bigl((i+i(a+iR))\dot(a+(a+iR)e^{it})\bigr)dt$$

which seems "ugly" so perhaps I am missing something.

If $\alpha\le 0$ we have $\vert e^{\alpha z}\vert \le 1$ in the "upper" plane, so that $\vert \int_{\gamma_R}\frac{\exp(\alpha\cdot z)}{z^2}dz\vert\le \frac{1}{(a+iR)^2}\cdot\pi R$ which gives us $$\int_{D_R}f(z)dz=\int_{a-iR}^{a+iR}f(x)dx+\int_{\Gamma_R}f(z)dz,$$

How can I continue ?

Answer 1

For technical reasons it is probably better to join $a+iR$ and $a-iR$ with an arc of a circle centered at the origin. If we denote with $\Gamma_R$ such arc, $|f(z)|=\left|\frac{\exp(\alpha z)}{z^2}\right|$ is bounded by $$ \frac{\exp(a|\alpha|)}{R^2+a^2} $$ over $\Gamma_R$, and the length of $\Gamma_R$ does not exceed $\pi\sqrt{R^2+a^2}$, so the contribute $$ \int_{\Gamma_R}\frac{\exp(\alpha z)}{z^2}\,dz $$ is negligible for large $R$ as wanted: only the residue at $z=0$ and the sign of $\alpha$ really matter.

Answer 2

The only pole of the integrand is at $0$, where it has residue $\alpha$.

For $\alpha \ge 0$, use the residue theorem with your contour $D_R$ (where the semicircle extends to the left) and note that $|\exp(\alpha z)| = \exp(\alpha \text{Re}(z)) \le \exp(\alpha a)$ for $z \in D_R$. Thus

$$\left|\int_{\Gamma_R} f(z) \; dz \right| \le \exp(\alpha a) \cdot \text{length}(\Gamma_R)/R^2 \to 0$$ Since $0$ is inside the contour for large $R$, the result will be the residue at $0$, namely $\alpha$.

For $\alpha \le 0$, make the semicircle extend to the right, and you get a similar bound. This time $0$ is not inside the contour, so the result will be $0$.