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

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.

  • $\begingroup$ using that the integrand is even, it should be even possible to apply the residue theorem to the first integral. $\endgroup$ – tired Jan 22 '15 at 16:22

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.


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

  • $\begingroup$ If you know that $F(x)=F(-x)$ then $\int_{-\infty}^{\infty}F(x)dx=2\int_{0}^{\infty}F(x)dx=$, then why you did the useless differentiation s to show tthat; you don't even provide some info regarding the first result, in case the OP don't know much bout it $\endgroup$ – RE60K Jan 22 '15 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.