$\int_{0}^{\frac{\pi}{2}} \frac{\cos \left(s \arctan \left(-\frac{x}{\ln \cos x}\right)\right)}{\left(x^{2}+\ln ^{2} \cos x\right)^{\frac{s}{2}}}d x$

Question

In the middle of solving for $\int_{0}^{\frac{\pi}{2}} \frac{\cos \left(s \arctan \left(-\frac{x}{\ln \cos x}\right)\right)}{\left(x^{2}+\ln ^{2} \cos x\right)^{\frac{s}{2}}} d x, 0<s<1$, I encountered the following substitution: $$\int_{0}^{\frac{\pi}{2}} \frac{\cos \left(\operatorname{sarctan}\left(-\frac{x}{\ln \cos x}\right)\right)}{\left(x^{2}+\ln ^{2} \cos x\right)^{\frac{5}{2}}} d x=\frac{1}{\Gamma(s)} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} u^{s-1} \cos (u x) e^{u \ln \cos x} d u d x$$ However, I'm not sure of how it is done. I tried \begin{aligned} &=\frac{1}{\Gamma(s)} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} t^{s-1} \cos \left(s \arctan \left(-\frac{x}{\ln \cos x}\right)\right) e^{-t \sqrt{x^{2}+\ln ^{2} \cos x}} d t d x \\ &=\frac{1}{\Gamma(s)} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} t^{s-1} \cos \left(s \arctan \left(-\frac{x}{\ln \cos x}\right)\right) e^{\frac{t \cos x}{\cos \left(\arctan \left(-\frac{x}{\ln \cos x}\right)\right)}} d t d x\end{aligned} But I still can't figure out how to transfer this to the ideal form. Please, help me out.

Answer 1

The integral can be written in the form

$$I=\int_{0}^{\pi/2}\frac{\cos(-s\arctan(\frac{x}{\ln\cos x}))}{(x^2+\ln^2\cos x)} =\Re\int_{0}^{\pi/2}(z(x))^{-s}dx$$

where we defined

$$z(x)=-\ln\cos x-ix$$

Now upon inserting the identity

$$\frac{1}{z^s}=\frac{1}{\Gamma(s)} \int_{0}^{\infty}du ~u^{s-1}e^{-zu}$$

we obtain that:

$$I=\frac{1}{\Gamma(s)}\Re\int_0^{\pi/2}dx\int_{0}^{\infty}du ~~u^{s-1}e^{u(\ln\cos x+ix)}=\frac{1}{\Gamma(s)}\int_0^{\pi/2}dx\int_{0}^{\infty}du ~~u^{s-1}e^{u\ln\cos x}\cos(ux)$$

as advertised.