To evaluate the integral using contour integration, consider the integral
$$\oint_C dz \frac{z^{-p} \log{(1+z)}}{1+z} $$
where $C$ is the following contour:
The magnitude of the integral about the large arc of radius $R$ behaves as $\frac{\log{R}}{R^p}$ as $R \to \infty$ and thus vanishes. Let the radius of the small circular arcs be $\epsilon$. The contour integral is then equal to, in this limit,
$$\left (1-e^{-i 2 \pi p} \right ) \int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} - e^{-i \pi p}\int_{\infty}^{1+\epsilon} dx \frac{x^{-p} [\log{(x-1)}+i \pi]}{1-x} \\ - e^{-i \pi p}\int_{1+\epsilon}^{\infty} dx \frac{x^{-p} [\log{(x-1)}-i \pi]}{1-x}+i \epsilon \int_{\pi}^{-\pi} d\phi \,e^{i \phi} \frac{(e^{i \pi}+\epsilon e^{i \phi})^{-p} \log{(\epsilon e^{i \phi})}}{\epsilon e^{i \phi}}$$
The first integral represents the integral about the branch cut along the positive real axis, which concerns the $z^{-p}$ term only. Note that the integral about the origin vanishes as $\epsilon \to 0$. The second and third integrals represent the integrals along each side of the branch cut on the negative axis. Note that, along this branch cut, the argument of $z^{-p}$ is $-\pi p$ on either side of the branch cut, as the branch cut there concerns the log term only. The fourth integral is the integral about the branch point $z=-1$.
By Cauchy's theorem, the contour integral is zero. Thus, we have
$$\left (1-e^{-i 2 \pi p} \right ) \int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = i 2 \pi \, e^{-i \pi p} \int_{1+\epsilon}^{\infty} dx \frac{x^{-p}}{x-1} + i e^{-i \pi p} \int_{-\pi}^{\pi} d\phi \left ( \log{\epsilon} + i \phi \right )$$
Note that
$$\begin{align}\int_{1+\epsilon}^{\infty} dx \frac{x^{-p}}{x-1} &= \int_0^{1-\epsilon} dx \frac{x^{p-1}}{1-x} + O(\epsilon) \end{align} $$
and
$$\begin{align}\int_0^{1-\epsilon} dx \frac{x^{p-1}}{1-x} &= \int_0^{1-\epsilon} dx \, x^p \left (\frac1x + \frac1{1-x} \right ) \\ &= \frac1p (1-\epsilon)^p - \left [\log{(1-x)} x^p \right ]_0^{1-\epsilon} + p \int_0^{1-\epsilon} dx \, x^{p-1} \log{(1-x)}\\ &= \frac1p - \log{\epsilon} + p \int_0^1 dx \, x^{p-1} \log{(1-x)} + O(\epsilon)\\ &= \frac1p - \log{\epsilon} - p \sum_{k=1}^{\infty} \frac1{k (k+p)}+ O(\epsilon) \\ &= \frac1p - \log{\epsilon}-\gamma -\psi(1+p)+ O(\epsilon) \\ &= - \log{\epsilon} - (\gamma + \psi(p))+ O(\epsilon) \end{align} $$
where $\psi$ is the digamma function. Note that the singular $\log{\epsilon}$ pieces cancel. The second piece of the second integral on the RHS vanishes as it is an odd function over a symmetric interval. Thus, we may take the limit as $\epsilon \to 0$ and we get
$$\int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = -\frac{i 2 \pi \, e^{-i \pi p}}{1-e^{-i 2 \pi p}} (\gamma + \psi(p)) = -\frac{\pi}{\sin{\pi p}} (\gamma + \psi(p))$$
Alternatively, we may express this so that it is clear that the integral takes a positive value:
$$\int_0^{\infty} dx \frac{x^{-p} \log{(1+x)}}{1+x} = \frac{\pi}{\sin{\pi p}} \left (\frac1p - H_p \right ) $$
where $H_p$ is the analytically continued harmonic number at $p$.
The main question is how do we want to define $\frac1{\sqrt{x^2-1}}$ along $[-1,1]$? There are two possibilities: $\frac1{\sqrt{x^2-1}}=\frac{\pm i}{\sqrt{1-x^2}}$. Once we decide that, things are pretty simple.
We can define
$$
\log\left(\frac{z+1}{z-1}\right)
=\log(3)+\int_2^z\left(\frac1{w+1}-\frac1{w-1}\right)\mathrm{d}w\tag{1}
$$
where the integral is evaluated along any path which does not intersect $[-1,1]$. A closed path avoiding $[-1,1]$ will circle both poles an equal number of times and the residues will cancel.
Therefore, $(1)$ defines $\log\left(\frac{z+1}{z-1}\right)$ with a branch cut along $[-1,1]$. Using $(1)$, we can define
$$
\frac1{\sqrt{z^2-1}}=\frac1{z+1}e^{\frac12\log\left(\frac{z+1}{z-1}\right)}\tag{2}
$$
According $(2)$, the integrand along the top of $[-1,1]$ is $\frac{-i}{\sqrt{1-z^2}}$ and along the bottom of $[-1,1]$ is $\frac{i}{\sqrt{1-z^2}}$. The integral around the two dumbbell ends vanish as their size gets smaller. Thus, the integral counter-clockwise along the whole dumbbell is
$$
4i\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}\tag{3}
$$
The integral of $\frac1{\sqrt{z^2-1} }$, as defined in $(2)$, counter-clockwise around a circle of essentially infinite radius is $2\pi i$.
Cauchy's Integral Theorem says that the integral around the dumbbell and a circle of essentially infinite radius are the same. Thus, $(3)$ says
$$
4i\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}=2\pi i\tag{4}
$$
We are back to the question I raised at the beginning: how to define $\frac1{\sqrt{x^2-1}}$ along $[-1,1]$.
If we look at $\frac1{\sqrt{x^2-1}}$ as $\frac1{\sqrt{z^2-1}}$ along the top of $[-1,1]$, then $\frac1{\sqrt{x^2-1}}=\frac{-i}{\sqrt{1-x^2}}$ and we get
$$
\int_0^1\frac{\mathrm{d}x}{\sqrt{x^2-1}}=-i\frac\pi2\tag{5}
$$
If we look at $\frac1{\sqrt{x^2-1}}$ as $\frac1{\sqrt{z^2-1}}$ along the bottom of $[-1,1]$, then $\frac1{\sqrt{x^2-1}}=\frac{i}{\sqrt{1-x^2}}$ and we get
$$
\int_0^1\frac{\mathrm{d}x}{\sqrt{x^2-1}}=i\frac\pi2\tag{6}
$$
Best Answer
For $t\equiv-\tau<0$, consider a half-circle of radius $M$ centred at $c$ and lying on the right of its diameter that goes from $c-iM$ to $c+iM$. By Cauchy's theorem $$ \int_{c-iM}^{x+iM}\frac{e^{tz}}{\sqrt{1+z^2}}dz =\int_{-\pi/2}^{+i\pi/2}\frac{e^{-\tau(c+Me^{i\varphi})}}{\sqrt{1+(c+Me^{i\varphi})^2}}iMe^{i\varphi}d\varphi; $$ the right-hand side is bounded in absolute value by an argument in the style of Jordan's lemma, and hence goes to $0$ as $M\to\infty$. So, indeed: $$\boxed{ \int_{c-i\infty}^{c+i\infty}\frac{e^{tz}}{\sqrt{1+z^2}}dz=0,\text{ for }t<0.} $$ For $t=0$ the integral diverges logarithmically, but we can compute its Cauchy principal value: $$ \int_{c-iM}^{x+iM}\frac{1}{\sqrt{1+z^2}}dz=\left[ \sinh^{-1}z \right]_{c-iM}^{c+iM}=\log\frac{cM^{-1}+i+\sqrt{(cM^{-1}+i)^2+M^{-2}}}{cM^{-1}-i+\sqrt{(cM^{-1}-i)^2+M^{-2}}} $$ and for $M\to\infty$ $$ \boxed{ PV\int_{c-i\infty}^{c+i\infty}\frac{1}{\sqrt{1+z^2}}dz=i\pi. } $$ Finally, for $t>0$, consider the contour below:
It is easy to see that the integrals along the horizontal segments vanish in the $M\to\infty$ limit, as well as the integral along the arc on the left (the latter, again by Jordan's lemma). Even the integrals on the small arcs give no contribution as the contour approaches the branch cuts.
The only contribution comes from the branch discontinuities: $$ \int_{c-i\infty}^{c+i\infty}\frac{e^{tz}}{\sqrt{1+z^2}}dz= 4i \int_{1}^{+\infty}\frac{{\sin (ty)}}{\sqrt{y^2-1}}dy. $$
Now, letting $y=\cosh \psi$, we have $$ 4i\Im \int_1^{+\infty}\frac{e^{ity}}{\sqrt{y^2-1}}dy= 2i\Im \int_{-\infty}^{+\infty}e^{it\cosh\psi}d\psi= 2i\Im \left(i\pi H_0^{(1)}(t)\right)=i2\pi J_0(t), $$ thanks to the integral representation of cylindrical Bessel functions. In this step, the analytic continuation $t\mapsto t+i\delta$, for a small $\delta>0$ which is then sent to $0$, has been employed. So, finally $$ \boxed{ \int_{c-i\infty}^{c+i\infty}\frac{e^{tz}}{\sqrt{1+z^2}}dz =i2\pi J_0(t), \text{ for }t>0. } $$ To sum up $$ f(t)= \begin{cases} 0 &\text{if }t<0\\ 1/2 &\text{if }t=0\text{ (in the }PV\text{ sense)}\\ J_0(t)&\text{if }t>0. \end{cases} $$