show that $\int_0^{\infty}\sin(u\cosh x)\sin(u\sinh x)\frac{dx}{\sinh x}=\frac{\pi }{2}\sin u$

Question

$$I(a)=\int_0^{\infty}\sin(u\cosh x)\sin(u\sinh x)\frac{dx}{\sinh x}:a>0$$

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.

Answer 1

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.

Answer 2

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 $$I_{1}+I_{2}=\int_{0}^{\infty}\dfrac{\cos{(ut)}}{1-t^2}dt=\dfrac{\pi}{2}\sin{u}$$