Gradshteyn&Ryzhik 3.976 states that $$\int_0^{\infty } \left(x^2+1\right)^{b-\frac{1}{2}} e^{-p x^2} \cos \left((2 b-1) \tan ^{-1}(x)+2 p x\right) \ dx=\frac{e^{-p} \sin (\pi b) \Gamma (b)}{2 p^b}$$ For $b,p>0$. This seems to be related to Hermite formula i.e. formula $7$ here, but so far not success. Any help or suggestions will be appreciated.


1 Answer 1


Notice that the integral in question $I$ can be written as


which we can subsequently manipulate into a gamma function integral by completing the square :


It is left to the reader to verify that since the integrand of the last expression has no poles in the complex plane, we can form a rectangular contour connecting the two lines of integration and prove that


Performing carefully a change of variables $y=x^2/p$ we find that


So we finally see that

$$I_1=-ie^{-p}e^{\pi ib}\frac{\Gamma(b)}{p^b}$$

and by extracting the real part and dividing by 2, we obtain the requested result:

$$I=e^{-p}\frac{\sin{(\pi b)}\Gamma(b)}{2p^b}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .