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I want to evaluate the following integral $$\int_0^\infty x^{t-1}e^{-\beta x}dx$$ where $\beta$ is a complex number.

Now, if $\beta$ was real, we could just set $y = \beta x$ and we will reduce to the Gamma function. Since $\beta$ is complex, though, when I set $y = \beta x$, I am integrating over the line with $\arg \beta$ on the complex plane, so I can't reduce directly to the Gamma function, can I?

I have found after some calculations that $$\int_0^\infty x^{t-1}e^{-\beta x}dx = \Gamma(t)\beta^{-t}$$ which is exactly what one would find if it didn't bother with the previous observation.

So my questions are: Is my observation on the complex line correct? and 2) What is the best way to prove the result?

My work

Write $\beta = a + ib$. Consider the integral as a function of $t,a,b$ to get $$I(t,a,b) = \int_0^\infty x^{t-1}e^{-a x}e^{-ibx}dx$$.

Notice that $$\frac{\partial I}{\partial a}(t,a,b) = -I(t+1,a,b)$$ and $$\frac{\partial I}{\partial b}(t,a,b) = -iI(t+1,a,b)$$

Now since $\displaystyle I(t+1) = \frac t{a+ib}I(t)$, the previous two equations become $$\frac{\partial I}{\partial a} = -\frac t{a+ib}I$$ and $$\frac{\partial I}{\partial b} = -\frac{it}{a+ib}I$$

which put together yield $I(t,a,b) = C(t) (a+ib)^{-t}$. Also, since $\displaystyle I(t,1,0) = C(t) = \int_0^\infty x^{t-1}e^{-x}dx= \Gamma(t)$, we get

$$I(t,a,b) = \Gamma(t) (a+ib)^{-t} = \Gamma(t)\beta^{-t}$$

which seems like too much work!

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    $\begingroup$ You can also first prove that the integral is analytic e.g. using these methods and then the result follows from analytic continuation being unique. $\endgroup$ – Count Iblis Aug 16 '15 at 20:21
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You're correct in the result. To verify this and to provide an instructive way forward, we present here a direct approach the relies on Cauchy's Integral Theorem.

To that end, we let $I$ be the integral in the complex plane given by

$$I=\oint_C z^{t-1}e^{-\beta z}\,dz$$

where $\text{Re}(\beta)>0$, $t>0$, and $C=C_1+C_2+C_3$ is the contour comprised of the $4$ components

$(i)$ $C_1$ is the real-line segment from $(\epsilon,0)$ to $(R,0)$;

$(ii)$ $C_2$ is the circular arc centered at the origin with radius $R$ from $(R,0)$ to $(R\cos \theta_0,R\sin \theta_0)$ where $\beta e^{i\theta_0}=1 \implies e^{i\theta_0}=\frac{1}{\beta}$.

$(iii)$ $C_3$ is the line segment from $(R\cos \theta_0,R\sin \theta_0)$ to $(\epsilon \cos \theta,\epsilon \sin \theta)$;

$(iv)$ $C_4$ is the circular arc centered at the origin with radius $\epsilon$ from $(\epsilon \cos \theta,\epsilon \sin \theta)$ to $(\epsilon,0)$.

Since $z^{x-1}e^{-\beta z}$ is analytic in $C$, then from Cauchy's Integral Theorem, we have $I=0$. Moreover, the contributions from $C_2$ and $C_4$ can easily be shown to vanish as $R\to \infty$ and $\epsilon \to 0$, respectively.

Therefore, we have

$$\begin{align} I&=\oint_C z^{t-1}e^{-\beta z}\,dz\\\\ &=\int_0^{\infty}(x)^{t-1}e^{-\beta x}\,dx+\int_\infty^0 (e^{i\theta_0}x)^{t-1}e^{-\beta e^{i\theta_0}x}e^{i\theta_0}dx=0\\\\ \implies \int_0^{\infty}(x)^{t-1}e^{-\beta x}\,dx&=\frac{1}{\beta^{t}}\int_0^\infty x^{t-1}e^{-x}\,dx\\\\ &=\frac{1}{\beta^{t}}\Gamma(t) \end{align}$$

which was to be shown!

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  • $\begingroup$ veri nice indeed. Thank you! :-) $\endgroup$ – Ant Aug 18 '15 at 9:18
  • $\begingroup$ @ant Thank you! Much appreciative. And you're welcome. My pleasure. $\endgroup$ – Mark Viola Aug 18 '15 at 14:21
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$$ f(\beta) = \int_0^\infty x^{t-1}e^{-\beta x}dx$$ is clearly an analytic function in $\beta$ if $\text{Re}(\beta) > 0$ (differentiate under the integral sign). $$ g(\beta) = \Gamma(t) \beta^{-t} $$ is also analytic in $\beta$. Since $f(\beta) = g(\beta)$ for all $\beta \in (0,\infty)$, it follows by the principle of isolated zeros that they must be equal on the intersections of their domains of definition.

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  • $\begingroup$ Analytic continuation is one of the coolest things ever. I still get excited whenever I see it in practice. $\endgroup$ – Cameron Williams Aug 17 '15 at 1:20
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    $\begingroup$ I was under the belief that the OP wanted to verify that $f=g$. $\endgroup$ – Mark Viola Aug 17 '15 at 1:30
  • $\begingroup$ @Dr.MV It is trivial by change of variables if $\beta$ is real, as the OP pointed out. I used analytic continuation to show it is true for all other $\beta$. $\endgroup$ – Stephen Montgomery-Smith Aug 17 '15 at 1:49
  • $\begingroup$ Your answer states that $f=g$ for all $\beta$, but how does one reach that conclusion immediately. Perhaps one can say $f=g$ for real $\beta>0$ and then use AC to conclude that $f=g$ for all $\beta$ with positive real part. $\endgroup$ – Mark Viola Aug 17 '15 at 1:55
  • $\begingroup$ @Dr.MV As it happens, the integral is undefined for the real part of $\beta$ less than or equal to 0. Nevertheless, it if were defined outside of that range, one could still use analytic continuation. $\endgroup$ – Stephen Montgomery-Smith Aug 17 '15 at 2:01

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