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An integral for the reciprocal Gamma function

As a preliminary to finding an integral representation of the Bessel function we shall first find an integral for the reciprocal Gamma function. Consider the integral

\begin{displaymath}I=\int_C (-t)^{z-1}e^{-t}dt,
\end{displaymath} (137)

where Rez>0 and where the contour C starts slightly above the real axis at $+\infty$, runs down to t=0, where it goes around counter-clockwise in a small circle and returns to $+\infty$ just below the real axis. The t-plane is cut from 0 to $\infty$. We define the logarithm such that $\ln (-t)$ is real on the negative axis. Thus, on the negative real axis arg(-t)=0. On the contour C we have $-\pi <{\rm arg} (-t)<\pi$, such that arg(-t)=0 on the negative real t-axis. Therefore, just above the positive real t-axis we have arg$(-t)=-\pi $, whereas just below we have arg$(-t)=+\pi$, the angle being measured counter-clockwise. It then follows that

\begin{displaymath}(-t)^{z-1}=e^{-i\pi (z-1)}t^{z-1}
\end{displaymath} (138)

just above the positive real axis, and

\begin{displaymath}(-t)^{z-1}=e^{+i\pi (z-1)}t^{z-1}
\end{displaymath} (139)

just below the positive real axis. On the small circle enclosing t=0 we have $-t=\delta e^{i\theta}$. Then
I = $\displaystyle \int_\infty^\delta e^{-i\pi (z-1)}t^{z-1}e^{-t}dt+\int_\delta^\in...
...^{i\theta})^{z-1}
e^{\delta (\cos\theta+i\sin\theta)}\delta e^{i\theta}id\theta$  
  $\textstyle \rightarrow$ $\displaystyle -2i\sin (\pi z)\int_0^\infty t^{z-1}e^{-t}dt
~{\rm for}~\delta\rightarrow 0.$ (140)

It was used that the integral over the small circle vanishes as $\delta^z$. The integral on the right hand side is the usual representation for the Gamma function, so $I=-2i\sin (\pi z)\Gamma(z)$, or

\begin{displaymath}\Gamma (z)=\frac{-1}{2i\sin (\pi z)}\int_C (-t)^{z-1}e^{-t}dt.
\end{displaymath} (141)

This is Hankel's formula for the Gamma function, valid for all $z\neq 0,
\pm 1,\pm 2,...$. Using the formula

 \begin{displaymath}\Gamma (z)\Gamma(1-z)=\frac{\pi}{\sin \pi z},
\end{displaymath} (142)

derived in the book in eqs. (3-49)-(3-50), we get the following integral representation for the reciprocal Gamma function,

 \begin{displaymath}\frac{1}{\Gamma (z)}=\frac {i}{2\pi}\int_\infty^{(0+)} (-t)^{-z}e^{-t}dt.
\end{displaymath} (143)

Here we have written $\int_\infty^{(0+)}$ for $\int_C$, meaning thereby a path starting at infinity on the real axis, encircling 0 in a positive sense, and returning to infinity along the real axis, respecting the cut along the positive real axis.


next up previous
Next: Integral representations of the Up: No Title Previous: A guide to the
Mette Lund
2000-04-27