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 $a$).


1 Answer 1


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.

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

  • $\begingroup$ Amazing. Thank you! I will try to look for more tricks like this in the future when I'm computing integrals. $\endgroup$
    – weirdmath
    Commented Sep 29, 2023 at 10:41
  • $\begingroup$ 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! $\endgroup$
    – weirdmath
    Commented Sep 29, 2023 at 10:55
  • 1
    $\begingroup$ @weirdmath That makes sense, good to hear! :) I'd recommend sticking to what you did then if that's the case. $\endgroup$
    – Frank W
    Commented Sep 29, 2023 at 18:12

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