Question 7.36: Evaluate 1/2πi ∫a-i∞^a+i∞ e^zt/√z + 1 dz where a and t are a...
Evaluate \frac{1}{2 \pi i} \int\limits_{a-i \infty}^{a+i \infty} \frac{e^{z t}}{\sqrt{z+1}} d z where a and t are any positive constants.
The integrand has a branch point at z=-1. We shall take as a branch line that part of the real axis to the left of z=-1. Since we cannot cross this branch line, let us consider
\oint\limits_{C} \frac{e^{z t}}{\sqrt{z+1}} d z
where C is the contour A B D E F G H J K A shown in Fig. 7-15. In this figure, E F and H J actually lie on the real axis but have been shown separated for visual purposes. Also, F G H is a circle of radius \epsilon while B D E and J K A represent arcs of a circle of radius R.
Since e^{z t} / \sqrt{z+1} is analytic inside and on C, we have by Cauchy’s theorem
\oint\limits_{C} \frac{e^{z t}}{\sqrt{z+1}} d z=0 (1)
Omitting the integrand, this can be written
\int\limits_{A B}+\int\limits_{B D E}+\int\limits_{E F}+\int\limits_{F G H}+\int\limits_{H J}+\int\limits_{J K A}=0 (2)
Now, on B D E and J K A, z=R e^{i \theta} where \theta goes from \theta_{0} to \pi and \pi to 2 \pi-\theta_{0}, respectively.
On E F, z+1=u e^{\pi i}, \sqrt{z+1}=\sqrt{u} e^{\pi i / 2}=i \sqrt{u}; whereas on H J, z+1=u e^{-\pi i}, \sqrt{z+1}=\sqrt{u} e^{-\pi i / 2}= -i \sqrt{u}. In both cases, z=-u-1, d z=-d u, where u varies from R-1 to \epsilon along E F and \epsilon to R-1 along H J.
On F G H, z+1=\epsilon e^{i \phi} where \phi goes from -\pi to \pi. Thus, (2) can be written
\begin{aligned} \int\limits_{a-i T}^{a+i T} \frac{e^{z t}}{\sqrt{z+1}} d z & +\int\limits_{\theta_{0}}^{\pi} \frac{e^{R e^{i \theta t}}}{\sqrt{R e^{i \theta}+1}} i R e^{i \theta} d \theta+\int\limits_{R-1}^{\epsilon} \frac{e^{-(u+1) t}(-d u)}{i \sqrt{u}} \\ & +\int\limits_{\pi}^{-\pi} \frac{e^{\left(\epsilon e^{i \phi}-1\right) t}}{\sqrt{\epsilon e^{i \phi}+1}} i \epsilon e^{i \phi} d \phi+\int\limits_{\epsilon}^{R-1} \frac{e^{-(u+1) t}(-d u)}{-i \sqrt{u}} \\ & +\int\limits_{\pi}^{2 \pi-\theta_{0}} \frac{e^{R e^{i \theta} t}}{\sqrt{R e^{i \theta}+1}} i R e^{i \theta} d \theta=0 & (3) \end{aligned}
Let us now take the limit as R \rightarrow \infty (and \left.T=\sqrt{R^{2}-a^{2}} \rightarrow \infty\right) and \epsilon \rightarrow 0. We can show (see Problem 7.111) that the second, fourth, and sixth integrals approach zero. Hence, we have
\int\limits_{a-i \infty}^{a+i \infty} \frac{e^{z t}}{\sqrt{z+1}} d z=\underset{\substack{\epsilon \rightarrow 0 \\ R \rightarrow \infty}}{\lim} 2 i \int\limits_{\epsilon}^{R-1} \frac{e^{-(n+1) t}}{\sqrt{u}} d u=2 i \int\limits_{0}^{\infty} \frac{e^{-(u+1) t}}{\sqrt{u}} d u
or letting u=v^{2}
\frac{1}{2 \pi i} \int\limits_{i \infty}^{a+i \infty} \frac{e^{z t}}{\sqrt{z+1}} d z=\frac{1}{\pi} \int\limits_{0}^{\infty} \frac{e^{-(u+1) t}}{\sqrt{u}} d u=\frac{2 e^{-t}}{\pi} \int\limits_{0}^{\infty} e^{-v^{2} t} d v=\frac{e^{-t}}{\sqrt{\pi t}}
