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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 $$\int e^{-ax}\sin(bx)=\frac{1}{a^2+b^2}e^{-ax}(-a\sin(bx)-b\cos(bx))$$ $$\int e^{-ax}\cos(bx)dx=\frac{1}{a^2+b^2}e^{-ax}(b\sin(bx)-a\cos(bx))$$

Chappers
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user5389726598465
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  • 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? – user5389726598465 Jun 03 '17 at 15:03

2 Answers2

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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 $\theta$.

user5389726598465
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Chappers
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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}

xpaul
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  • On the last line, how do you get $\Re\frac{1}{(a^2+b^2-i2ab)^{m/2}}=\frac{1}{(a^2+b^2)^{m/2}}\cos(m\tan^{-1}(\frac{b}{a}))$? – user5389726598465 Jun 03 '17 at 15:32
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    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. – Mark Viola Jun 03 '17 at 15:42
  • @MarkViola, you are right and I just omit the detail. – xpaul Jun 03 '17 at 18:01
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    @user135711, $a-bi=r(\cos\theta-i\sin\theta)$ where $r=\sqrt{a^2+b^2}$, $\theta=\tan^{-1}(\frac{b}{a})$. – xpaul Jun 03 '17 at 18:03
  • @xpaul Yes, and I'm suggesting you include it. – Mark Viola Jun 03 '17 at 18:11
  • @MarkViola, ok. – xpaul Jun 03 '17 at 18:23