$$ I = \int_{0}^{\infty} \frac{\sqrt{2} \left( e^{-2vz - 2b\sinh(z)}\cos\left(\pi\left(\frac{1}{4} + v\right)\right) + e^{2vz - 2b\sinh(z)}\sin\left(\pi\left(\frac{1}{4} + v\right)\right) \right)}{\sqrt{e^z - e^{-z}}} \, \mathrm{d}z $$ Going through Desmos, the function does converge.$f(z)$, the integrand, is holomorphic in its domain, and the limits at $+\infty$ and $-\infty$ give me zero as long as $v \in \mathbb{R}$ and $\text{Re}(b) > 0$.
Edit:
As FShrike had mentioned, there's a branch cut at $0$. This is the reason for the keyhole contour. For further context, see the comments.
I wanted to also first show the Proof of Convergence from Desmos and two interesting cases of the integral. 
$$\\$$
Case 1:
If I have $I(b)$ equal to the integral and vary $v$, I get close similarities to the Modified Bessel Function of the Second Kind.
$$\\$$
Case 2: If I have $I(v)$ equal to the integral, and vary $b$, the graph and its properites can be seen below.
$\\$
I also wanted to provide an alternative contour that might work as well. After realization, there's a branch cut at $z=i\pi$ as well.
Also, as I tested equivalencies for the bounds of the integral, I got these odd results:
Note:
This might not be necessary but may aid in a solution, here's how the integral behaves under various values of $v$ and $b$
Question:
If stated correctly, the contour integral evaluates to 0 due to Cauchy's Integral Theorem. However, the path integrals of the contour would have to be evaluated. How would you evaluate the path integrals and evaluate the integral with the given notes above?
$$
\oint_{c} \frac{{\sqrt{2} \times (e^{-2vz - 2b\sinh(z)}\cos(\pi(1/4 + v)) + e^{2vz - 2b\sinh(z)}\sin(\pi(1/4 + v)))}}{{\sqrt{e^z - e^{-z}}}} \, dz
$$
Using the contour
$$ \oint_c = \lim_{{R \to \infty, \epsilon \to 0}} \left( \int_{\Gamma} + \int_{-R}^{-\epsilon} + \int_{\psi} + \int_{\epsilon}^{R} \right) $$ And with what I had stated: $$ 0 =\lim_{{R \to \infty, \epsilon \to 0}} \left( \int_{\Gamma} + \int_{-R}^{-\epsilon} + \int_{\psi} + \int_{\epsilon}^{R} \right) $$ Please excuse my drawing as I am not an artist, Keyhole Contour:
Progress Made $$ \lim_{{R \to \infty \atop \epsilon \to 0}} \left(\int_{-R}^{-\epsilon} + \int_{\epsilon}^{R}\right)$$ $\\$ Which is equal to: $\int_{-\infty}^{0} + I$. And in connection to my notes: $I=-\int_{-\infty}^{0} f(z) dz$. Now, things get messy, so I'll keep it concise. $\\$ Evaluating: $\int_{-\infty}^{0} f(z) \, dz$ $\\$ First I made the substitution $ w = e^{z},dw = e^{z} \, dz $. $$\int_{-\infty}^{\infty} \frac{e^{b(1/w-w)}w^{-1-2v}\left(\cos \left(\pi \left(\frac{1}{4}+v\right)\right)+w^{4v}\sin \left(\pi \left(\frac{1}{4}+v\right)\right) \right)}{\sqrt{-\frac{1}{w}+w}} dw $$ Distribute the $\pi$, use the sum and difference formulas for $\cos$ and $\sin$, and distribute throughout. Next I factor out $\frac{\sqrt{2}}{2}$ which cancels out At this point, the integral is: $$ \int_{-\infty}^{\infty} \frac{\left(e^{-bw + \frac{b}{w}} w^{(-1 - 2v)}\cos(\pi v) - e^{-bw + \frac{b}{w}} w^{(2v - 1)}\cos(\pi v)\right) + \left(-e^{-bw + \frac{b}{w}} w^{(-1 - 2v)}\sin(\pi v) + e^{-bw + \frac{b}{w}} w^{(2v - 1)}\sin(\pi v)\right)}{\sqrt{w - w^{-1}}} \, dw$$ Split into two fractions, $$I_{1} = \int_{-\infty}^{\infty} \frac{\cos(\pi v) e^{-bw + \frac{b}{w}} (w^{-1 - 2v} - w^{2v - 1})}{\sqrt{w - w^{-1}}} dw$$ And, $$I_{2}=\int_{-\infty}^{\infty}\frac{\sin\left(\pi v\right)e^{-bw+\frac{b}{w}}\left(w^{\left(-1-2v\right)}+w^{\left(2v-1\right)}\right)}{\sqrt{w-w^{-1}}}dw$$ Therefore, $I=-(I_1+I_2)$ I have also assumed both path integrals evaluate to zero. However, I'm uncertain.