# §13.4 Integral Representations

## §13.4(i) Integrals Along the Real Line

 13.4.1 ${\mathbf{M}}\left(a,b,z\right)=\frac{1}{\Gamma\left(a\right)\Gamma\left(b-a% \right)}\int_{0}^{1}e^{zt}t^{a-1}(1-t)^{b-a-1}\mathrm{d}t,$ $\Re b>\Re a>0$, ⓘ Symbols: $\Gamma\left(\NVar{z}\right)$: gamma function, ${\mathbf{M}}\left(\NVar{a},\NVar{b},\NVar{z}\right)$: Olver’s confluent hypergeometric function, $\mathrm{d}\NVar{x}$: differential, $\mathrm{e}$: base of natural logarithm, $\int$: integral, $\Re$: real part and $z$: complex variable A&S Ref: 13.2.1 (in different form) Referenced by: §13.10(v), §13.29(iii) Permalink: http://dlmf.nist.gov/13.4.E1 Encodings: TeX, pMML, png See also: Annotations for §13.4(i), §13.4 and Ch.13
 13.4.2 ${\mathbf{M}}\left(a,b,z\right)=\frac{1}{\Gamma\left(b-c\right)}\int_{0}^{1}{% \mathbf{M}}\left(a,c,zt\right)t^{c-1}(1-t)^{b-c-1}\mathrm{d}t,$ $\Re b>\Re c>0$,
 13.4.3 ${\mathbf{M}}\left(a,b,-z\right)=\frac{z^{\frac{1}{2}-\frac{1}{2}b}}{\Gamma% \left(a\right)}\int_{0}^{\infty}e^{-t}t^{a-\frac{1}{2}b-\frac{1}{2}}J_{b-1}% \left(2\sqrt{zt}\right)\mathrm{d}t,$ $\Re a>0$.

For the function $J_{b-1}$ see §10.2(ii).

 13.4.4 $U\left(a,b,z\right)=\frac{1}{\Gamma\left(a\right)}\int_{0}^{\infty}e^{-zt}t^{a% -1}(1+t)^{b-a-1}\mathrm{d}t,$ $\Re a>0$, $|\operatorname{ph}{z}|<\frac{1}{2}\pi$,
 13.4.5 $U\left(a,b,z\right)=\frac{z^{1-a}}{\Gamma\left(a\right)\Gamma\left(1+a-b\right% )}\int_{0}^{\infty}\frac{U\left(b-a,b,t\right)e^{-t}t^{a-1}}{t+z}\mathrm{d}t,$ $|\operatorname{ph}{z}|<\pi$, $\Re a>\max\left(\Re b-1,0\right)$,
 13.4.6 $U\left(a,b,z\right)=\frac{(-1)^{n}z^{1-b-n}}{\Gamma\left(1+a-b\right)}\int_{0}% ^{\infty}\frac{{\mathbf{M}}\left(b-a,b,t\right)e^{-t}t^{b+n-1}}{t+z}\mathrm{d}t,$ $\left|\operatorname{ph}z\right|<\pi$, $n=0,1,2,\dots$, $-\Re b,
 13.4.7 $U\left(a,b,z\right)=\frac{2z^{\frac{1}{2}-\frac{1}{2}b}}{\Gamma\left(a\right)% \Gamma\left(a-b+1\right)}\*\int_{0}^{\infty}e^{-t}t^{a-\frac{1}{2}b-\frac{1}{2% }}K_{b-1}\left(2\sqrt{zt}\right)\mathrm{d}t,$ $\Re a>\max\left(\Re b-1,0\right)$,
 13.4.8 $U\left(a,b,z\right)=z^{c-a}\*\int_{0}^{\infty}e^{-zt}t^{c-1}{{}_{2}{\mathbf{F}% }_{1}}\left(a,a-b+1;c;-t\right)\mathrm{d}t,$ $|\operatorname{ph}{z}|<\frac{1}{2}\pi$,

where $c$ is arbitrary, $\Re c>0$. For the functions $K_{b-1}$ and ${{}_{2}{\mathbf{F}}_{1}}$ see §10.25(ii) and §§15.1, 15.2(i).

## §13.4(ii) Contour Integrals

 13.4.9 ${\mathbf{M}}\left(a,b,z\right)=\frac{\Gamma\left(1+a-b\right)}{2\pi\mathrm{i}% \Gamma\left(a\right)}\int_{0}^{(1+)}e^{zt}t^{a-1}{(t-1)^{b-a-1}}\mathrm{d}t,$ $b-a\neq 1,2,3,\dots$, $\Re a>0$.
 13.4.10 ${\mathbf{M}}\left(a,b,z\right)=e^{-a\pi\mathrm{i}}\frac{\Gamma\left(1-a\right)% }{2\pi\mathrm{i}\Gamma\left(b-a\right)}\int_{1}^{(0+)}e^{zt}t^{a-1}{(1-t)^{b-a% -1}}\mathrm{d}t,$ $a\neq 1,2,3,\dots$, $\Re\left(b-a\right)>0$. Figure 13.4.1: Contour of integration in (13.4.11). (Compare Figure 5.12.3.) Magnify
 13.4.11 ${\mathbf{M}}\left(a,b,z\right)=e^{-b\pi\mathrm{i}}\Gamma\left(1-a\right)\Gamma% \left(1+a-b\right)\*\frac{1}{4\pi^{2}}\int_{\alpha}^{(0+,1+,0-,1-)}e^{zt}t^{a-% 1}{(1-t)^{b-a-1}}\mathrm{d}t,$ $a,b-a\neq 1,2,3,\dots$.

The contour of integration starts and terminates at a point $\alpha$ on the real axis between $0$ and $1$. It encircles $t=0$ and $t=1$ once in the positive sense, and then once in the negative sense. See Figure 13.4.1. The fractional powers are continuous and assume their principal values at $t=\alpha$. Similar conventions also apply to the remaining integrals in this subsection.

 13.4.12 ${\mathbf{M}}\left(a,c,z\right)=\frac{\Gamma\left(b\right)}{2\pi\mathrm{i}}z^{1% -b}\int_{-\infty}^{(0+,1+)}e^{zt}t^{-b}{{}_{2}{\mathbf{F}}_{1}}\left(a,b;c;% \ifrac{1}{t}\right)\mathrm{d}t,$ $b\neq 0,-1,-2,\dots$, $\left|\operatorname{ph}z\right|<\frac{1}{2}\pi$.

At the point where the contour crosses the interval $(1,\infty)$, $t^{-b}$ and the ${{}_{2}{\mathbf{F}}_{1}}$ function assume their principal values; compare §§15.1 and 15.2(i). A special case is

 13.4.13 ${\mathbf{M}}\left(a,b,z\right)=\frac{z^{1-b}}{2\pi\mathrm{i}}\int_{-\infty}^{(% 0+,1+)}e^{zt}t^{-b}\!\left(1-\frac{1}{t}\right)^{-a}\mathrm{d}t,$ $|\operatorname{ph}{z}|<\frac{1}{2}\pi$.
 13.4.14 $U\left(a,b,z\right)=e^{-a\pi\mathrm{i}}\frac{\Gamma\left(1-a\right)}{2\pi% \mathrm{i}}\int_{\infty}^{(0+)}e^{-zt}t^{a-1}{(1+t)^{b-a-1}}\mathrm{d}t,$ $a\neq 1,2,3,\dots$, $\left|\operatorname{ph}z\right|<\frac{1}{2}\pi$.

The contour cuts the real axis between $-1$ and $0$. At this point the fractional powers are determined by $\operatorname{ph}{t}=\pi$ and $\operatorname{ph}\left(1+t\right)=0$.

 13.4.15 $\frac{U\left(a,b,z\right)}{\Gamma\left(c\right)\Gamma\left(c-b+1\right)}=\frac% {z^{1-c}}{2\pi\mathrm{i}}\int_{-\infty}^{(0+)}e^{zt}t^{-c}{{}_{2}{\mathbf{F}}_% {1}}\left(a,c;a+c-b+1;1-\frac{1}{t}\right)\mathrm{d}t,$ $\left|\operatorname{ph}z\right|<\frac{1}{2}\pi$.

Again, $t^{-c}$ and the ${{}_{2}{\mathbf{F}}_{1}}$ function assume their principal values where the contour intersects the positive real axis.

## §13.4(iii) Mellin–Barnes Integrals

If $a\neq 0,-1,-2,\dots$, then

 13.4.16 ${\mathbf{M}}\left(a,b,-z\right)=\frac{1}{2\pi\mathrm{i}\Gamma\left(a\right)}% \int_{-\mathrm{i}\infty}^{\mathrm{i}\infty}\frac{\Gamma\left(a+t\right)\Gamma% \left(-t\right)}{\Gamma\left(b+t\right)}z^{t}\mathrm{d}t,$ $|\operatorname{ph}{z}|<\tfrac{1}{2}\pi$,

where the contour of integration separates the poles of $\Gamma\left(a+t\right)$ from those of $\Gamma\left(-t\right)$.

If $a$ and $a-b+1\neq 0,-1,-2,\dots$, then

 13.4.17 $U\left(a,b,z\right)=\frac{z^{-a}}{2\pi\mathrm{i}}\int_{-\mathrm{i}\infty}^{% \mathrm{i}\infty}\frac{\Gamma\left(a+t\right)\Gamma\left(1+a-b+t\right)\Gamma% \left(-t\right)}{\Gamma\left(a\right)\Gamma\left(1+a-b\right)}z^{-t}\mathrm{d}t,$ $|\operatorname{ph}{z}|<\tfrac{3}{2}\pi$,

where the contour of integration separates the poles of $\Gamma\left(a+t\right)\Gamma\left(1+a-b+t\right)$ from those of $\Gamma\left(-t\right)$.

 13.4.18 $U\left(a,b,z\right)=\frac{z^{1-b}e^{z}}{2\pi\mathrm{i}}\int_{-\mathrm{i}\infty% }^{\mathrm{i}\infty}\frac{\Gamma\left(b-1+t\right)\Gamma\left(t\right)}{\Gamma% \left(a+t\right)}z^{-t}\mathrm{d}t,$ $|\operatorname{ph}{z}|<\tfrac{1}{2}\pi$,

where the contour of integration passes all the poles of $\Gamma\left(b-1+t\right)\Gamma\left(t\right)$ on the right-hand side.