Page:Deux Mémoires de Henri Poincaré.djvu/12

by Planck's theory. I did not stop at these calculations and I pass immediately to the principal question, whether the discontinuities that I just mentioned must necessarily be admitted.

I will reproduce the reasoning of Poincaré, but I will at first say that in the formulas that we will encounter, α indicates a complex variable of which the real part $\alpha _{r}$ is always positive. In the representation we will limit ourselves to the half of plane α characterized by $\alpha _{r}>0$, and in integrations in respect to α we will follow a straight line l perpendicular to the axis of real α, and prolonged indefinitely on the two sides. The values of the integrals will be independent of the length of the distance $\alpha _{r}>0$ of this line at the origin of α.

Poincaré introduced an auxiliary function that defines the equation

(11)	$\Phi (\alpha )=\int _{0}^{\infty }\omega (\eta )e^{-\alpha \eta }d\eta$

and demonstrated that the function ω and the derived function $\varphi$ can be be expressed by using Φ.

We obtain at first, by inverting (11)

(12)	$\omega (\eta )={\frac {1}{2i\pi }}\int _{(L)}\Phi (\alpha )e^{\alpha \eta }d\alpha$

For a similar formula for $\varphi (x)$ we notice that in equation (11) we can replace η by any of the variables $\eta _{1},\dots ,\eta _{n}$. Multiplying the n equations which we obtained, we find

$\left[\Phi (\alpha )\right]^{n}=\int _{0}^{\infty }\dots \int _{0}^{\infty }\omega \left(\eta _{1}\right)\dots \omega \left(\eta _{n}\right)e^{-\alpha x}d\eta _{1}\dots d\eta _{n}$

or, by virtue of the formula (8)

$\left[\Phi (\alpha )\right]^{n}=\int _{0}^{\infty }\varphi (x)^{-\alpha x}dx$

and by inversion

$\varphi (x)={\frac {1}{2\pi i}}\int _{(L)}\left[\Phi (\alpha )\right]^{n}e^{\alpha x}d\alpha$

The formulas (9) and (10) now become

${\begin{array}{l}nY={\frac {C}{2i\pi }}\int _{0}^{h}\int _{(L)}x(h-x)^{p-1}[\Phi (\alpha )]^{n}e^{\alpha x}dx\ d\alpha \\\\pX={\frac {C}{2i\pi }}\int _{0}^{h}\int _{(L)}(h-x)^{p}[\Phi (\alpha )]^{n}e^{\alpha x}dx\ d\alpha \end{array}}$