I am trying to solve the following integral $$ \displaystyle\int^{x}_0 f(t)\cos\alpha(x-t)\displaystyle\int^{t}_0 f(s)\cos\alpha(t-2s)ds dt $$ $\alpha$ is a increasing real sequence and the function $f$ is real and $f \in L_1$.
My idea is to use the subtraction formula for cosine, and then separate the integral into 4 cases:
$\cos(\alpha x)\displaystyle\int^{x}_0 f(t)\cos^2(\alpha t)\displaystyle\int^{t}_0 f(s)\cos (2\alpha s) ds dt$
$\cos(\alpha x)\displaystyle\int^{x}_0 f(t)\cos(\alpha t)\sin(\alpha t)\displaystyle\int^{t}_0 f(s)\sin (2\alpha s) ds dt$
$\sin(\alpha x)\displaystyle\int^{x}_0 f(t)\cos(\alpha t)\sin(\alpha t)\displaystyle\int^{t}_0 f(s)\cos (2\alpha s) ds dt$
$\sin(\alpha x)\displaystyle\int^{x}_0 f(t)\sin^2(\alpha t)\displaystyle\int^{t}_0 f(s)\sin (2\alpha s) ds dt$
Since $f \in L_1$, I want to use the property of Riemann-Lebesgue, so in the first integral:
- $\frac{\cos(\alpha x)}{2}\displaystyle\int^{x}_0 f(t)(1+\cos(2\alpha s))\displaystyle\int^{t}_0 f(s)\cos (2\alpha s) ds dt$
we can observe two things:
$\displaystyle\int^{x}_0 f(t)\displaystyle\int^{t}_0 f(s)\cos (2\alpha s) ds dt \rightarrow 0$, because $\displaystyle\int^{t}_0 f(s)\cos (2\alpha s) ds \rightarrow 0$ using Riemann-Lebesgue.
$\displaystyle\int^{x}_0 f(t)\cos(2\alpha s)\displaystyle\int^{t}_0 f(s)\cos (2\alpha s) ds dt = \displaystyle\int^{x}_0 Q(t)Q'(t) dt $, where $Q(t) =\displaystyle\int^{t}_0 f(s)\cos (2\alpha s) ds$ and with that $Q \rightarrow 0$ using Riemann-Lebesgue.
The second integral is similar, using that $2\cos(\alpha s)\sin(\alpha s)=\sin(2\alpha s)$
but the problem is in the last two integrals, where I have something like $\displaystyle\int^{x}_0 f(t)\cos(2\alpha s)\displaystyle\int^{t}_0 f(s)\sin (2\alpha s) ds dt$.
I would like to receive your help, to know how to attack this problem. Am I fine or is there another method? Perhaps doing a change of variable or using a product of cosines in the original integral.
thank you very much for your time and help