# Proof of: $\int_0^\infty x^{m-1}e^{-ax} \cos bx \ dx = \frac{\Gamma(m)}{(a^{2} + b^{2})^{m/2}}\cos\left(m\tan^{-1}\left(\frac{b}{a}\right)\right)$

Where can I find a proof or how do you prove the following: $$\int_0^\infty x^{m-1}e^{-ax} \cos bx \ dx = \frac{\Gamma(m)}{(a^{2} + b^{2})^{m/2}}\cos\left(m\tan^{-1}\left(\frac{b}{a}\right)\right)$$

Edit: I think I see the identity now For a cascade of integration by parts let $$\int e^{-ax}\sin(bx)=\frac{1}{a^2+b^2}e^{-ax}(-a\sin(bx)-b\cos(bx))$$ $$\int e^{-ax}\cos(bx)dx=\frac{1}{a^2+b^2}e^{-ax}(b\sin(bx)-a\cos(bx))$$

• Well, everyone knows (m-1)! and integrating by parts gives $\int e^{-ax}cos(bx)dx=\frac{1}{a^2+b^2}e^{-ax}(bsin(bx)-acos(bx)$. Does it just follow from nested integrating by parts? Jun 3, 2017 at 15:03

Here's how Euler did this (from E675, translation available here). We start with $$\int_0^{\infty} x^{m-1} e^{-x} \, dx = \Gamma(m).$$ Changing variables gives $$\int_0^{\infty} x^{m-1} e^{-kx} \, dx = \frac{\Gamma(m)}{k^m}.$$

Euler now assumes that this still works if $k=p \pm iq$ is complex, provided $p>0$. (It does, but this needs an application of Cauchy's theorem.) We then have $$\int_0^{\infty} x^{m-1} e^{-(p \pm iq)x} \, dx = \frac{\Gamma(m)}{(p \pm iq)^m},$$ and if we write $p=f\cos{\theta}$, $q=f\sin{\theta}$, we can apply Euler's formula to obtain $$\int_0^{\infty} x^{m-1} e^{-(p \pm iq)x} \, dx = \frac{\Gamma(m)}{f^m}(\cos{m\theta} \mp i\sin{m\theta}).$$ Taking the real part gives us $$\int_0^{\infty} x^{m-1} e^{-px} \cos{qx} \, dx = \frac{\Gamma(m)}{f^m}\cos{m\theta},$$ and the result follows by inverting the expressions of $p$ and $q$ in terms of $f$ and $\theta$.

• (This answer is taken in part from this other answer of mine.) Jun 3, 2017 at 15:47
• (+1) ... I had recommended the mentioning of the importance of CIT when deforming the contour. So, well done. Jun 3, 2017 at 18:37

Recalling $$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx$$ one has \begin{eqnarray} &&\int_0^\infty x^{m-1}e^{-ax} \cos bx \; dx\\ &=&\Re\int_0^\infty x^{m-1}e^{-ax} e^{ibx} \; dx\\ &=&\Re\int_0^\infty x^{m-1}e^{-(a-ib)x} \; dx\\ &=&\Re\frac{1}{(a-ib)^{m}}\int_0^\infty x^{m-1}e^{x} \; dx\\ &=&\Re\frac{1}{(a-ib)^{m}}\Gamma(m)\\ &=&\frac{1}{(a^2+b^2)^{m/2}}\Gamma(m)\cos(m\tan^{-1}(\frac{b}{a}))\\ \end{eqnarray}

• On the last line, how do you get $\Re\frac{1}{(a^2+b^2-i2ab)^{m/2}}=\frac{1}{(a^2+b^2)^{m/2}}\cos(m\tan^{-1}(\frac{b}{a}))$? Jun 3, 2017 at 15:32
• In enforcing the substitution $(a-ib)x\to x$, the upper limit of integration changes to $(a-bi)\infty$, which is not on the real axis in general. You need invoke Cauchy's Integral Theorem to deform the integration contour to the real line. You should consider mentioning this as it is a critical step that is often glossed over. Jun 3, 2017 at 15:42
• @MarkViola, you are right and I just omit the detail. Jun 3, 2017 at 18:01
• @user135711, $a-bi=r(\cos\theta-i\sin\theta)$ where $r=\sqrt{a^2+b^2}$, $\theta=\tan^{-1}(\frac{b}{a})$. Jun 3, 2017 at 18:03
• @xpaul Yes, and I'm suggesting you include it. Jun 3, 2017 at 18:11