show that $\int_0^{\infty}\sin(u\cosh x)\sin(u\sinh x)\frac{dx}{\sinh x}=\frac{\pi }{2}\sin u$ $$I(a)=\int_0^{\infty}\sin(u\cosh x)\sin(u\sinh x)\frac{dx}{\sinh x}:a>0$$
I started with $$\sin(a)\sin(b)=\frac{1}{2}(\cos(a-b)-\cos(a+b))$$
so $$I(a)=\frac{1}{2}\int_0^{\infty}\left ( \cos(ue^{-x})-\cos(ue^x) \right )\frac{dx}{\sinh x}$$
$t=e^x$
$$I(a)=\int_1^{\infty}\left ( \cos(ut^{-1})-\cos(ut) \right )\frac{dt}{t^2-1}$$
$t \to1/v$
$$I(a)=\int_0^{1}\left ( \cos(uv)-\cos(uv^{-1}) \right )\frac{dv}{1-v^2}$$
$$I(a)=\sum_{n=0}^{\infty}\int_0^{1}\left ( v^{2n}\cos(uv)-v^{2n}\cos(uv^{-1}) \right )dv$$
and I can't solve the last integral ---
so what is your suggestions to solve the last integral or if there is better way to start using real analysis.
 A: By the residue theorem,
$$ \int_{0}^{+\infty}\frac{\cos(uv)-\cos u}{1-v^2}\,dv = \frac{\pi}{2}(\operatorname{sign} u)\sin u.$$
Just consider the previous integral, split the integration range into $(0,1)\cup(1,+\infty)$ and use the substitution $v\to 1/v$ on the second interval.
A: It is well know this following result:
$$F(a)=\int_{-\infty}^{\infty}\dfrac{\cos{ax}}{1-x^2}dx=\pi\sin{a},a>0$$
because
$$\Longrightarrow F'(a)=\int_{-\infty}^{\infty}\dfrac{-x\sin{(ax)}}{1-x^2}dx
\Longrightarrow F''(a)=\int_{-\infty}^{+\infty}\dfrac{-x^2\sin{(ax)}}{1-x^2}dx$$
so we have $$\Longrightarrow F''(a)+F(a)=0,\Longrightarrow F(x)=C_{1}\cos{a}+C_{2}\sin{a}$$
Note  $$F(0)=0,F(0^{+})-F(0^{-})=-\pi$$
so we have
$$F(a)=\pi\sin{a}$$
so
$$\int_{0}^{\infty}\dfrac{\cos{(ax)}}{1-x^2}dx=\dfrac{\pi}{2}\sin{a}$$
since you have
$$I(u)=\int_{1}^{\infty}\dfrac{\cos{ut^{-1}}}{t^2-1}-\int_{1}^{\infty}\dfrac{\cos{ut}}{t^2-1}=I_{1}-I_{2}$$
and note
$$I_{1}=\int_{0}^{1}\dfrac{\cos{ut}}{1-t^2}$$
so
$$I_{1}+I_{2}=\int_{0}^{\infty}\dfrac{\cos{(ut)}}{1-t^2}dt=\dfrac{\pi}{2}\sin{u}$$
