Integral computation of $\int_0^\pi \mathrm d t \sin(a\cos t/2) \mathrm{sinh}(b\sin t/2)$ I'm having trouble computing an integral.
$$
I=\int_0^1 \frac{\mathrm{d}x}{2x(1-x)}\left(x-\cosh\left(\frac{t\sqrt{1-x}}{\tau}\right)+\sqrt{1-x}\text{ }\mathrm{sinh}\left(\frac{t\sqrt{1-x}}{\tau}\right)\right)\left(1-\cos\left(\frac{y\sqrt{x}}{2u\tau}\right)\right)
$$
I tried several changes of variable, such as $x=\cos^2(\frac{\theta}{2})$:
$$
I=\int_0^\pi \frac{\mathrm{d}\theta}{\sin(\theta)}
\left(
    \underbrace{\cos^2(\theta)-1}_{A_1}+
    \underbrace{1-\cosh\left(
        \frac{t}{\tau}\sin(\theta/2)
        \right)}_{A_2}+
    \underbrace{\sin\frac{\theta}{2}
    \text{ }\mathrm{sinh}\left(
        \frac{t}{\tau}\sin(\theta/2)
        \right)}_{A_3}
    \right)
\left(
    1-
    \cos\left(
        \frac{y}{2u\tau}\cos(\theta/2)\right)
        \right)
$$
The first term $A_1$ is computable, result is something with cosintegral, and is not a problem. One can consider $\frac{\mathrm d A_3}{\mathrm d y}$ and one gets:
$$
\frac{\mathrm d A_3}{\mathrm d y}=\frac{1}{4u\tau}
\int_0^\pi\mathrm d \theta \text{ } \mathrm{sinh}\left(
        \frac{t}{\tau}\sin(\theta/2)
        \right)
\sin\left(
        \frac{y}{2u\tau}\cos(\theta/2)\right)
$$
The $A_2$ term can be transform in something very similar:
$$
\frac{\mathrm d^2 A_3}{\mathrm d t\mathrm d y}=-
\frac{1}{4u\tau^2}
\underline{\int_0^\pi\mathrm d \theta \text{ } \mathrm{sinh}\left(
        \frac{t}{\tau}\sin(\theta/2)
        \right)
\sin\left(
        \frac{y}{2u\tau}\cos(\theta/2)\right)}_{\Large\text{this one}}
$$
Has anyone any idea how to compute such an integral ? I tried looking in the handbook for mathematical functions, and found that integral representation of Kelvin functions are almost the same as my integral. I am looking for any special (or not) function that may allow me to have an analytic expression of the latter integral.
 A: One way to get an expression for the integral, although not in closed form, is to use the Jacobi-Anger expansion:
$${{\rm e}^{iz\cos \left( \theta \right) }}=\sum _{n=-\infty }^{\infty }
{i}^{n}{{\rm J}_n\left(z\right)}{{\rm e}^{in\theta}} \tag{1}$$
to expand the integrand in terms of Bessel functions of the first kind ($J_n$). In so doing, it can be shown that:
$$\sinh \left( {\frac {t\sin \left( \frac{\theta}{2} \right) }{\tau}}
 \right) \sin \left( {\frac {y\cos \left( \frac{\theta}{2} \right) }{2u\tau}}
 \right) =\sum_{n=-\infty}^{\infty} \left( \sum _{k=-\infty}^{\infty} i\left( -1 \right) 
^{k}{{\rm J}_{2k+1}\left({\frac {y}{2u\tau}}\right)}
{{\rm J}_{2n+1}\left({\frac {-it}{\tau}}\right)}\sin \left( \frac{\,
 \left( 2\,n+1 \right) \theta}{2} \right) \cos \left(  \frac{\,
 \left( 2\,k+1 \right) \theta}{2} \right)  \right) \tag{2}$$
which can be written:
$$\frac{i}{2}\sum _{n=-\infty}^{\infty} \left( \sum _{k=-\infty}^{\infty}\left( -1
 \right) ^{k}{{\rm J}_{2k+1}\left({\frac {y}{2u\tau}}\right)} \left( 
{{\rm J}_{-2k-1+2n}\left({\frac {-it}{\tau}}\right)}+
{{\rm J}_{2k+1+2n}\left({\frac {-it}{\tau}}\right)} \right) \sin
 \left( n\theta \right)  \right) \tag{3}$$
and thus:

$$\int _{0}^{\pi }\!\sinh \left( {\frac {t\sin \left( 1/2\,\theta
 \right) }{\tau}} \right) \sin \left( {\frac {y\cos \left( 1/2\,\theta
 \right) }{2u\tau}} \right) {d\theta}=\sum _{n=-\infty}^{\infty} \left( \sum _{k=-\infty}^{\infty}\frac{i\left( -1
 \right) ^{k}{{\rm J}_{2k+1}\left({\frac {y}{2u\tau}}\right)} \left( 
{{\rm J}_{-2k+1+4n}\left({\frac {-it}{\tau}}\right)}+
{{\rm J}_{2k+3+4n}\left({\frac {-it}{\tau}}\right)} \right) }{2n+1}  \right) \tag{4}$$

It may not look too pretty but the series often converges very rapidly and it may simplify further. Convolutions of Bessel functions are sometimes known as generalised Bessel functions although this may be a stretch in terms of analytic functions.
Update
This way is probably neater and simpler. Note that:
$$\sinh \left( {\frac {t\sin \left( \frac{\theta}{2} \right) }{\tau}}
 \right) \sin \left( {\frac {y\cos \left( \frac{\theta}{2} \right) }{2u
\tau}} \right) =-\frac{i}{2} \left( \cos \left( x\sin \left( \frac{\theta}{2}+
\phi \right)  \right) -\cos \left( x\sin \left( \frac{\theta}{2}-\phi
 \right)  \right)  \right) \tag{i}$$
$$\phi=i\mathrm{arctanh} \left( {\frac {y}{2tu}} \right) \in \mathbb{C},\quad x=\frac{it}{2\tau}{\sqrt{4-{\frac {{y}^{2}}{{t}^{2}{u}^{2}}}}} \in \mathbb{C} \tag{ii} $$
where standard product-to-sum trig identities were used and then the two trig functions inside, each with the same argument, are written as one scaled and shifted trig function.
Then starting from $(1)$ you can prove:
$$-\frac{i}{2}\left[\cos \left( x\sin \left( \frac{\theta}{2}+
\phi \right)  \right) -\cos \left( x\sin \left( \frac{\theta}{2}-\phi
 \right)  \right)\right] =\\
i\sum _{n=-\infty }^{\infty }{{\rm J}_n\left(x\right)}\sin \left( n\frac{\theta}{2}\right) \sin \left( n\phi \right) \tag{iii}$$
integrate and simplify to get:

$$\int _{0}^{\pi }\!-\frac{i}{2}\left[\cos \left( x\sin \left( \frac{\theta}{2}+
\phi \right)  \right) -\cos \left( x\sin \left( \frac{\theta}{2}-\phi
 \right)  \right)\right] {d\theta}=\\4\,i\sum _{n=1}^{\infty }{\frac {
{{\rm J}_{4n-2}\left(x\right)}\sin \left( \left( 4n-2 \right)\, 
\phi \right) }{2\,n-1}} \tag{iv}$$

which checks out for all numerical values I have tried so far. Note that the integrals on the left individually look like Bessel functions $(J_0)$ but for the $\phi$ shift.
