# $\Gamma (x) \cos(ax)$ identity

I am asked to show that $$\Gamma (x) \cos(ax) = b^x \int_{0}^{\infty} \mathrm{d} t \enspace t^{x-1} e^{-bt \cos(a)} \cos(bt \sin(a)).$$

A change of variables $$t \to \frac{t}{b}$$ shows that $$b$$ is superfluous and we can just eliminate it from the RHS. The exercise recommends that I write $$\sin$$ and $$\cos$$ using Euler's formula, and so the RHS becomes

$$\frac{1}{2} \int_{0}^{\infty} \mathrm{d}t \enspace t^{x-1} e^{-t \cos(a)} \left[ e^{it \sin(a)} + e^{-it \sin(a)} \right]$$

and I want to get this to the form $$\frac{1}{2} \int_{0}^{\infty} \mathrm{d}t \enspace t^{x-1} e^{-t} \left[ e^{iax} + e^{-iax} \right] = \frac{1}{2} \int_{0}^{\infty} \mathrm{d}t \enspace t^{x-1} e^{-t \cos(a)} \left[ e^{iax + t \cos (a) - t} + e^{-iax + t \cos(a) -t} \right]$$

I have tried various things like performing $$t \to \frac{t}{\cos(a)}$$, $$t \to t \tan(a)$$, $$t \to \frac{t}{\sin(a)}$$. I have also tried integrating by parts the RHS (where I integrate $$\cos \exp$$) with no success. It seems there's some clever substitution that I'm just missing. I would appreciate any hints/guidance. Also, per Mathematica, the result I'm trying to show is indeed correct (given some constraints on $$a$$).

The trick in this case is to rewrite the cosine term as the real part of a single exponential function.

$$\cos(bt\sin a)=\Re(e^{ibt\sin a})$$

The integral becomes

$$\int_0^{+\infty}\mathrm dt\, t^{x-1}e^{-bt(\cos a-i\sin a)}=\int_0^{+\infty}\mathrm dt\, t^{x-1}\exp(-bte^{-ia})=b^{-x}e^{iax}\Gamma(x)$$

Considering only the real part, then $$\Re(e^{iax})=\cos ax$$, completing the proof.

$$\int_0^{+\infty}\mathrm dt\, t^{x-1} e^{-bt\cos a}\cos(bt\sin a)\color{blue}{=b^{-x}\Gamma(x)\cos ax}$$

• Amazing. Thank you! I will try to look for more tricks like this in the future when I'm computing integrals. Commented Sep 29, 2023 at 10:41
• Actually, I do realize now that the trick I was missing was combining sin and cos in the exponents. It does work if I do that in my 2nd expression, though considering only the real part is still nicer! Commented Sep 29, 2023 at 10:55
• @weirdmath That makes sense, good to hear! :) I'd recommend sticking to what you did then if that's the case. Commented Sep 29, 2023 at 18:12