Where did I go wrong with this Bromwich integral?

Question

I am trying to evaluate $$ \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{e^{pt}}{\sqrt{p+1}}\,dp. $$ I started by constructing a branch cut along $p=-1$ to $p=-\infty$ and set \begin{align*} \oint_C\tilde{f}(p)\,dp = \int_{c-i\infty}^{c+i\infty}\frac{e^{pt}}{\sqrt{p+1}}\,dp\ &+\ iR\int_{\pi/2}^\pi \frac{e^{tRe^{i\phi}}}{\sqrt{Re^{i\phi}+1}}\,e^{i\phi}d\phi\ +\ \int_{-\infty}^{-1}\frac{e^{pt}}{e^{i\pi/2}\sqrt{p+1}}\,dp\\ &+i\epsilon \int_{\pi}^{-\pi}\frac{e^{t\epsilon e^{i\theta }}}{\sqrt{\epsilon e^{i\theta}+1}}e^{i\theta}\,d\theta\ +\ \int_{-1}^{-\infty}\frac{e^{pt}}{e^{-i\pi/2}\sqrt{p+1}}\,dp. \end{align*} By Cauchy's theorem the integral on the LHS is $0$. The integrals over the arcs vanish and we find $$ 0= 2\pi i I+2i\int_{-1}^{-\infty}\frac{e^{pt}}{\sqrt{p+1}}\,dp. $$ In the end I got something with an imaginary number which is obvious based on the integration limits. So what was it that went wrong here? What I think I did wrong was integrating over $-\infty$ to $-1$, but I am not entirely sure why that is incorrect. The Bromwich contour closes in the left half plane so how could I integrate from $-1$ to $\infty$? Any help would be very much appreciated. Thank you!

Answer 1

Note that $\sqrt{p+1}$ is purely imaginary when $p\in (-\infty, -1)$. So on the upper part of the branch cut, $\sqrt{p+1}=i\sqrt{|p+1|}$ while on the lower part of the branch cut $\sqrt{p+1}=-i\sqrt{|p+1|}$. Then, we have

$$\begin{align} 0&=\int_{c-i\infty}^{c+i\infty}\frac{e^{pt}}{\sqrt{p+1}}\,dp+\int_{-\infty}^{-1}\frac{e^{pt}}{i\sqrt{|p+1|}}\,dp-\int_{-\infty}^{-1}\frac{e^{pt}}{-i\sqrt{|p+1|}}\,dp\\\\ &=\int_{c-i\infty}^{c+i\infty}\frac{e^{pt}}{\sqrt{p+1}}\,dp-2i\int_1^\infty \frac{e^{-pt}}{\sqrt{p-1}}\,dp\tag1 \end{align}$$

Now, we can easily evaluate the second integral on the right-hand side of $(1)$ by enforcing the substitution $p\mapsto 1+p^2$. Proceeding we find that

$$\int_1^\infty \frac{e^{-pt}}{\sqrt{p-1}}\,dp=2\int_0^\infty e^{-tp^2}\,dp=\frac{2\sqrt \pi e^{-t}}{\sqrt t}$$

And now you can finish.