Estimating integral along arc in the contour integral $\int_{c}z^{a-1}e^{-c z}\,dz$

Question

I came across the following integral in Bateman´s Higher Transcendental Functions (Vol I, first chapter equation 32)

\begin{align*} \int_0^\infty t^{a-1} e^{-c t \cos \beta-ic t \sin \beta }\,dt=\frac{e^{-ia \beta}\Gamma(a)}{c^a} \tag{1} \end{align*}

For $-\pi/2< \beta < \pi/2, \,\, \text{Re}(a)>0$ or $\beta= \pm \pi/2, \,\,0< \text{Re}(a)<1.$ To proof $(1)$ the author considers the contour integral

\begin{align*} \int_{c}z^{a-1}e^{-c z}\,dz \qquad \text{for} \,\,c>0 \end{align*}

Integrated around the contour (and it´s mirrored image around the real axis) below

Since $f(z)$ is analytic inside this contour, by Cauchy´s theorem:

\begin{align*} \int_{c}z^{a-1}e^{-c z}\,dz=0 \end{align*}

Writing all integrals explicitly we obtain

\begin{align*} & \int_\epsilon^R x^{a-1}e^{- c x}\,dx +i\int_0^{\beta}\left(R e^{i \theta} \right)^{a-1}e^{-c R e^{i \theta}}R e^{i \theta}\,d\theta \\ & \qquad - e^{i \beta}\int_\epsilon^R \left(t e^{i \beta} \right)^{a-1} e^{-c t e^{i \beta}}\,dt-i \int_0^\beta \left(R e^{i \phi} \right)^{a-1}e^{-c R e^{i \phi}}R e^{i \phi}\,d\phi\\ &=I_1+I_2+I_3+I_4=0 \end{align*}

Upon letting $\epsilon \to 0$ and $R \to \infty$

\begin{align*} & \int_0^\infty x^{a-1}e^{- c x}\,dx - e^{i \beta}\int_0^\infty t^{a-1} e^{-c t \cos \beta-ic t \sin \beta }\,dt=0 \tag{2} \end{align*}

which gives the desired result.

I am strugling to proof that the two integrals around the arcs vanish when $\epsilon \to 0$ and $R \to \infty$.

For the integral along the big arc I obtained

\begin{align*} \left| i\int_0^{\beta}\left(R e^{i \theta} \right)^{a-1}e^{-c R e^{i \theta}}R e^{i \theta}\,d\theta \right| &\leq R^{\text{Re}(a)} \int_0^{\beta} e^{-c R \cos \theta- \text{Im}(a) \theta}\,d \theta\\ &=R^{\text{Re}(a)} \int_0^{\beta} e^{-c R \sin\left(\frac{\pi}{2}- \theta\right)- \text{Im}(a) \theta}\,d \theta\\ \end{align*}

similarly, for the integral along the small arc

\begin{align*} \left| i\int_0^{\beta}\left(\epsilon e^{i \phi} \right)^{a-1}e^{-c \epsilon e^{i \phi}}R e^{i \phi}\,d\phi \right| &\leq \epsilon^{\text{Re}(a)} \int_0^{\beta} e^{-c \epsilon \cos \phi- \text{Im}(a) \phi}\,d \phi\\ \end{align*}

If the integrand was of the form $e^{-c R \sin \theta}$ we could use the inequality $e^{-c R \sin \theta} \leq e^{-c 2 R \theta/\pi}$ for $0 \leq \theta \leq \pi/2$ to estimate the integral., but this is not the case here.

Answer 1

After some thought I came up with the following estimation for the integral along the big arc

\begin{align*} \left| i\int_0^{\beta}\left(R e^{i \theta} \right)^{a-1}e^{-c R e^{i \theta}}R e^{i \theta}\,d\theta \right| &\leq R^{\text{Re}(a)} \int_0^{\beta} e^{-c R \cos \theta- \text{Im}(a) \theta}\,d \theta\\ &\leq R^{\text{Re}(a)} \int_0^{\beta} e^{-c R \left(1-2 \theta/\pi \right)- \text{Im}(a) \theta}\,d \theta\\ &= R^{\text{Re}(a)}e^{-c R}\int_0^{\beta} e^{\left(2 c R /\pi - \text{Im}(a)\right) \theta}\,d \theta\\ &= \frac{\pi R^{\text{Re}(a)-1}e^{-c R}}{2c- \frac{\pi\text{Im}(a)}{R}}\left(e^{\left(\frac{2 c R }{\pi} - \text{Im}(a)\right) \beta} -1\right)\\ &= \frac{\pi R^{\text{Re}(a)-1}}{2c- \frac{\pi\text{Im}(a)}{R}}\left(e^{ - \text{Im}(a) \beta} e^{-cR\left(1-\frac{2 \beta}{\pi} \right)} - e^{-c R}\right)\\ \end{align*}

and this vanishes as $R \to \infty$ provided $-\pi/2< \beta < \pi/2, \,\, \text{Re}(a)>0$ or $\beta= \pm \pi/2, \,\,0< \text{Re}(a)<1.$

Answer 2

The integral around the smaller arc in $\epsilon$ vanishes as $\epsilon\to0$. To see that you just need to observe that: $$\lim_{z\to0}z\cdot f(z)=\lim_{z\to0}z^a\cdot e^{-cz}=0$$Since $\Re a>0$, where $f$ denotes the integrand.

Likewise, the integral around the larger arc vanishes, simply because: $$\lim_{R\to\infty}\sup_{|\phi|\le|\beta|}R\cdot(R^{a-1}\cdot|e^{-c Re^{i\phi}}|)=\lim_{R\to\infty}R^a\cdot e^{-R\cdot c\cos\beta}=0$$Then, use the very trivial ML lemma to conclude: $$\left|\oint_\gamma f\right|\le\ell(\gamma)\cdot\sup_{z\in\gamma}|f(z)|$$