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I have the following function defined piecewise continuous with $S,T\in\mathbb{R} > 0$ $$f(x)=\begin{cases} \frac{e^{T/u}}{u^S} & x < 0 \\ 0 & x \geq 0 \end{cases} $$

and I want to normalize this function to integrate to $1$. This creates a particularly tricky integral though that I have never seen before and I was wondering if anyone knew about how to get the value in terms of $S,T$

$$ \int_{-\infty}^0 \frac{e^{T/x}}{x^S}dx $$

I've tried the standard methods, but this appears harder then I'm used to, if anyone has some insight!

wjmccann
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2 Answers2

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This integral is very close to the integral definition of the gamma function which is $$\Gamma(s) = \int_{0}^{\infty} x^{s-1} e^{-x} dx$$

First use the substitution $u = \frac{1}{x} , -\frac{1}{u^2} du = dx$ to get

$$\int_{-\infty}^{0} \frac{e^{T/x}}{x^s} dx = \int_{-\infty}^{0} e^{Tu} u^{s-2} du$$

(As we are approaching 0 from the left, $\lim_{x \rightarrow 0^{-}} = -\infty$)

Next use the substitution $v = -Tu, \frac{-dv}{T} = du$ to get

$$(-1)^s T^{1-s} \int_{0}^{\infty} e^{-v} v^{s-2} dv = (-1)^s T^{1-s} \Gamma(s-1)$$

Arahat
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  • This makes sense, but it appears that this integral could provide both a complex, and a negative result, when the function as definied is real and positive over the entire domain – wjmccann Aug 04 '20 at 13:44
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This seems to be an example of the incomplete gamma function.

The (upper) incomplete gamma function is defined as $\Gamma(s,x) = \int_x^{\infty} t^{s-1}e^{-t} dt$

Now, $\int_{-\infty}^{0}\frac{e^{T/x}}{x^S} dx = \int_{-\infty}^{\infty}\frac{e^{T/x}}{x^S}dx - \int_{0}^{\infty}\frac{e^{T/x}}{x^S}dx$

You should now be able to bring the terms in the standard form to use the incomplete gamma function using variable substitution

John White
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