Integral of hyperbolic and ordinary trigonometric function Recently I have proved that
\begin{equation}
\int_{-\infty}^{\infty}dx\frac{\cos\left[\lambda x\right]}{\cosh^{2}x}=\frac{\pi\lambda}{\sinh\left[\pi\lambda/2\right]}
\end{equation}
Now I would like to get an integral result for a slight more complicated formula, given by
\begin{equation}
\int_{-\infty}^{\infty}dx\frac{\cos\left[\lambda x\right] \cos[\Lambda x^2]}{\cosh^{2}x}
\end{equation}
Do you know any tricks or hints for where I should start? Or does this equation have an analytical result? I have tried looking within integral tables, but I could not find this integral in particular. Mathematica also did not give me any answer.
Thanks!
 A: If written cleverly, this has a very surprising connection to quantum physics. Let's denote
$$I[\Lambda,\lambda] = \int_{-\infty}^\infty \frac{\cos[\lambda x]\exp[i\Lambda x^2]}{\cosh^2x}dx$$
where the integral we want is $\operatorname{Re}\{I[\Lambda,\lambda]\}$. Taking derivatives, we have
$$\partial_\Lambda I = \int_{-\infty}^\infty \frac{ix^2\cos[\lambda x]\exp[i\Lambda x^2]}{\cosh^2x}dx$$
$$\partial_\lambda^2I = \int_{-\infty}^\infty \frac{-x^2\cos[\lambda x]\exp[i\Lambda x^2]}{\cosh^2x}dx$$
$$\implies i\partial_\Lambda I = \partial_\lambda^2 I$$
In other words, $I$ is a solution to the Schrodinger equation (for a $1$D free particle) where $\Lambda$ plays the role of a time variable. This has a known solution in the form of the time evolution operator $U[\Lambda]$ for a time-independent Hamiltonian
$$I[\Lambda,\lambda] = U[\Lambda]I[0,\lambda] = \exp\left[-i\Lambda \partial_\lambda^2\right]\frac{\pi \lambda}{\sinh\left[\frac{\pi \lambda}{2}\right]}$$
We could go even further and simplify this expression for our integral
$$\operatorname{Re}\{I[\Lambda,\lambda]\} = \cos\left[\Lambda \partial_\lambda^2\right]\frac{\pi \lambda}{\sinh\left[\frac{\pi \lambda}{2}\right]}$$
meaning the first term to even appear as a correction would be a fourth derivative of the expression.
