# §13.4 Integral Representations

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

 13.4.1 $\mathop{{\mathbf{M}}\/}\nolimits\!\left(a,b,z\right)=\frac{1}{\mathop{\Gamma\/% }\nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits\!\left(b-a\right)}\int_{0% }^{1}e^{zt}t^{a-1}(1-t)^{b-a-1}\mathrm{d}t,$ $\Re{b}>\Re{a}>0$,
 13.4.2 $\mathop{{\mathbf{M}}\/}\nolimits\!\left(a,b,z\right)=\frac{1}{\mathop{\Gamma\/% }\nolimits\!\left(b-c\right)}\int_{0}^{1}\mathop{{\mathbf{M}}\/}\nolimits\!% \left(a,c,zt\right)t^{c-1}(1-t)^{b-c-1}\mathrm{d}t,$ $\Re{b}>\Re{c}>0$,
 13.4.3 $\mathop{{\mathbf{M}}\/}\nolimits\!\left(a,b,-z\right)=\frac{z^{\frac{1}{2}-% \frac{1}{2}b}}{\mathop{\Gamma\/}\nolimits\!\left(a\right)}\int_{0}^{\infty}e^{% -t}t^{a-\frac{1}{2}b-\frac{1}{2}}\mathop{J_{b-1}\/}\nolimits\!\left(2\sqrt{zt}% \right)\mathrm{d}t,$ $\Re{a}>0$.

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

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

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

## §13.4(ii) Contour Integrals

 13.4.9 $\mathop{{\mathbf{M}}\/}\nolimits\!\left(a,b,z\right)=\frac{\mathop{\Gamma\/}% \nolimits\!\left(1+a-b\right)}{2\pi\mathrm{i}\mathop{\Gamma\/}\nolimits\!\left% (a\right)}\int_{0}^{(1+)}e^{zt}t^{a-1}{(t-1)^{b-a-1}}\mathrm{d}t,$ $b-a\neq 1,2,3,\dots$, $\Re{a}>0$.
 13.4.10 $\mathop{{\mathbf{M}}\/}\nolimits\!\left(a,b,z\right)=e^{-a\pi\mathrm{i}}\frac{% \mathop{\Gamma\/}\nolimits\!\left(1-a\right)}{2\pi\mathrm{i}\mathop{\Gamma\/}% \nolimits\!\left(b-a\right)}\int_{1}^{(0+)}e^{zt}t^{a-1}{(1-t)^{b-a-1}}\mathrm% {d}t,$ $a\neq 1,2,3,\dots$, $\Re{(b-a)}>0$.
 13.4.11 $\mathop{{\mathbf{M}}\/}\nolimits\!\left(a,b,z\right)=e^{-b\pi\mathrm{i}}% \mathop{\Gamma\/}\nolimits\!\left(1-a\right)\mathop{\Gamma\/}\nolimits\!\left(% 1+a-b\right)\*\frac{1}{4\pi^{2}}\int_{\alpha}^{(0+,1+,0-,1-)}e^{zt}t^{a-1}{(1-% t)^{b-a-1}}\mathrm{d}t,$ $a,b-a\neq 1,2,3,\dots$.

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

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

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

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

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

Again, $t^{-c}$ and the $\mathop{{{}_{2}{\mathbf{F}}_{1}}\/}\nolimits$ 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 $\mathop{{\mathbf{M}}\/}\nolimits\!\left(a,b,-z\right)=\frac{1}{2\pi\mathrm{i}% \mathop{\Gamma\/}\nolimits\!\left(a\right)}\int_{-\mathrm{i}\infty}^{\mathrm{i% }\infty}\frac{\mathop{\Gamma\/}\nolimits\!\left(a+t\right)\mathop{\Gamma\/}% \nolimits\!\left(-t\right)}{\mathop{\Gamma\/}\nolimits\!\left(b+t\right)}z^{t}% \mathrm{d}t,$ $|\mathop{\mathrm{ph}\/}\nolimits{z}|<\tfrac{1}{2}\pi$,

where the contour of integration separates the poles of $\mathop{\Gamma\/}\nolimits\!\left(a+t\right)$ from those of $\mathop{\Gamma\/}\nolimits\!\left(-t\right)$.

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

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

where the contour of integration separates the poles of $\mathop{\Gamma\/}\nolimits\!\left(a+t\right)\mathop{\Gamma\/}\nolimits\!\left(% 1+a-b+t\right)$ from those of $\mathop{\Gamma\/}\nolimits\!\left(-t\right)$.

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

where the contour of integration passes all the poles of $\mathop{\Gamma\/}\nolimits\!\left(b-1+t\right)\mathop{\Gamma\/}\nolimits\!% \left(t\right)$ on the right-hand side.