How to evaluate these two integrals about hyperbolic functions? While I was calculating the two integrals below
\begin{align*}
\mathcal{I}&=\int_{0}^{\infty }\frac{\cos x}{1+\cosh x-\sinh x}\mathrm{d}x\\
\mathcal{J}&=\int_{0}^{\infty }\frac{\sin x}{1+\cosh x-\sinh x}\mathrm{d}x
\end{align*}
I used series expansion,and the two integrals became the series below
$$\mathcal{I}=\sum_{n=0}^{\infty }\frac{n\cdot\left ( -1 \right )^{n}}{1+n^{2}}~,~\mathcal{J}=\sum_{n=0}^{\infty }\frac{ \left ( -1 \right )^{n}}{1+n^{2}}$$
but I don't know how to evaluate the two series,the WolframAlpha gave me the answer about digamma function:
\begin{align*}
\mathcal{I}&=\frac{1}{4}\left [ \psi ^{\left ( 0 \right )}\left ( \frac{1}{2}-\frac{i}{2} \right )+\psi ^{\left ( 0 \right )}\left ( \frac{1}{2}+\frac{i}{2} \right )-\psi ^{\left ( 0 \right )}\left ( 1-\frac{i}{2} \right )-\psi ^{\left ( 0 \right )}\left ( 1+\frac{i}{2} \right ) \right ]\\
\mathcal{J}&=\frac{i}{4}\left [ -\psi ^{\left ( 0 \right )}\left ( \frac{1}{2}-\frac{i}{2} \right )+\psi ^{\left ( 0 \right )}\left ( \frac{1}{2}+\frac{i}{2} \right )+\psi ^{\left ( 0 \right )}\left ( 1-\frac{i}{2} \right )-\psi ^{\left ( 0 \right )}\left ( 1+\frac{i}{2} \right ) \right ]
\end{align*}
I'd like to know is there a simple form for it or how to evaluate the two series.Without using series expansion,how to solve the two integrals.
EDIT:
I found a way to calculate the integral $\mathcal{J}$,I will post it later.
 A: \begin{align*}
\mathcal{J}=\int_{0}^{\infty }\frac{\sin x}{1+\cosh x-\sinh x}\mathrm{d} x&=\int_{0}^{\infty } \frac{\sin x}{1+e^{-x}}\mathrm{d}x\\
&=\sum_{n=0}^{\infty }\left ( -1 \right )^{n}\int_{0}^{\infty }e^{-nx}\sin x\mathrm{d}x=\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{1+n^{2}}\\
&=\sum_{n=0}^{\infty }\frac{1}{\left ( 2n \right )^{2}+1}-\sum_{n=0}^{\infty }\frac{1}{\left ( 2n +1\right )^{2}+1}\\
&=\frac{1}{2}+\frac{\pi }{4}\coth \left ( \frac{\pi }{2} \right )-\frac{\pi }{4}\tanh\left ( \frac{\pi }{2} \right )\\
&=\frac{1}{2}+\frac{\pi }{2}\mathrm{csch} \pi 
\end{align*}
A: $$1+\cosh x +\sinh x=1+e^{-x}$$
$$I+i J=\int_0^{+\infty} {e^{ix} \over 1+e^{-x}} dx$$
Update
As mentioned by user mickep this integration does not converge. The denominator is converging to 1 and the numerator is fluctuating.
A: 
Wolfram Alpha gave me the answer about digamma function: $$\begin{align*}
\mathcal{I}&=\frac{1}{4}\left[+~\psi^{\left(0\right)}\left(\frac{1}{2}-\frac{i}{2}\right)+\psi^{\left(0\right)}\left(\frac{1}{2}+\frac{i}{2}\right)-\psi^{\left(0\right)}\left(1-\frac{i}{2}\right)-\psi^{\left(0\right)}\left(1+\frac{i}{2}\right)\right]\\
\mathcal{J}&=\frac{i}{4}\left[-~\psi^{\left(0\right)}\left(\frac{1}{2}-\frac{i}{2}\right)+\psi^{\left(0\right)}\left(\frac{1}{2}+\frac{i}{2}\right)+\psi^{\left(0\right)}\left(1-\frac{i}{2}\right)-\psi^{\left(0\right)}\left(1+\frac{i}{2}\right)\right]
\end{align*}$$ But Mathematica $9$ shows $\displaystyle\mathcal{J}=\frac{1}{2}(1+\pi\text{csch}(\pi)).$

Hint: Do you notice the recurrent expressions of the form $\psi_0(1\pm x)$ and $\psi_0\bigg(\dfrac12\pm x\bigg)$ ?  Now, are you aware of the reflection formula for the digamma and polygamma function,  which can be deduced by differentiating the natural logarithm of Euler's infinite product  formula$($s$)$ for the $($co$)$sine function ?
A: 
The picture shows that the integral isn't divergent.
