In [his 1926 paper](https://ipparco.roma1.infn.it/pagine/deposito/2011/TF.pdf) Fermi states without further explanation that it follows from the Thomas-Fermi equation
$$\frac{d^2\psi(x)}{dx^2}=\frac{\psi(x)^{3/2}}{\sqrt{x}},\label{1} \tag{1}$$
and boundary conditions
$$\psi(0)=1,\quad\psi(\infty)=0\label{2}\tag{2},$$
that
$$\int\limits\_0^\infty{\frac{\psi^{5/2}(x)}{\sqrt{x}}}dx=-\frac{5}{7}\psi^\prime(0),\label{3} \tag{3}$$ where $\psi^\prime(x)=\frac{d\psi(x)}{dx}$. It seems Fermi considered this inference trivial. However, the proof given below and inspired by [Kleinert, p. 429](https://hagenkleinert.de/documents/pi/HagenKleinert_PathIntegrals.pdf), although simple, is not at all trivial.


The Thomas-Fermi equation \eqref{1} can be considered as a Euler-Lagrange equation coresponding to the action principle
$$
\delta S=0,\;\; S=\int\limits\_0^\infty\left [\frac{1}{2}\left (\frac{d\psi}{dx}\right)^2+\frac{2}{5}\frac{\psi^{5/2}(x)}{\sqrt{x}}\right ]dx.\label{4}\tag{4}
$$
The following infinitesimal deformation of the "coordinate" function
$$\psi(x)\to \bar\psi(x)=\psi(\lambda x),\;\lambda=1+\epsilon,\;\epsilon\ll 1,$$
respects the boundary conditions \eqref{2} and changes the action functional \eqref{4} to
$$\bar S=\int\limits\_0^\infty\left [\frac{1}{2}\left (\frac{d\bar\psi(x)}{dx}\right)^2+\frac{2}{5}\frac{\bar\psi^{5/2}(x)}{\sqrt{x}}\right ]dx=\int\limits\_0^\infty\left [\frac{\lambda}{2}\left (\frac{d\psi(y)}{dy}\right)^2+\frac{2}{5}\frac{\psi^{5/2}(y)}{\sqrt{\lambda y}}\right ]dy,$$ where $y=\lambda x$. Therefore, using $\lambda =1+\epsilon,\;\lambda^{-1/2}\approx 1-\epsilon/2$, we get
$$\bar S =S+\epsilon \int\limits\_0^\infty\left [\frac{1}{2}\left (\frac{d\psi(y)}{dy}\right)^2-\frac{1}{5}\frac{\psi^{5/2}(y)}{\sqrt{y}}\right ]dy,$$
and since we must have $\delta S=\bar S-S=0$ for any $\epsilon\ll 1$, we conclude
$$\int\limits\_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx=\frac{2}{5}\int\limits\_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}} dx.\label{5} \tag{5}$$
On the other hand, using \eqref{1}, \eqref{2} and integration by parts, we have
$$
\begin{split}
\int\limits\_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}} dx &=\int\limits\_0^\infty \psi(x)\frac{d^2\psi(x)}{dx^2}dx\\
&=\left .\psi(x)\frac{d\psi(x)}{dx}\right |\_0^\infty-\int\limits\_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx\\
& =-\left .\frac{d\psi(x)}{dx}\right |\_{x=0}-\int\limits\_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx,
\end{split}$$ and the Fermi result \eqref{3} immediately follows from \eqref{5}.


Is there a simpler proof of \eqref{3} that Fermi might have had in mind?



How did Fermi calculate this integral?
The way this is taught in text books,$^\ast$ which is likely the way Fermi reasoned, is to compare two alternative integrations by parts. One the one hand,
$$\int\_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}}\,dx=-5\int\_0^\infty \psi'(x)\psi^{3/2}(x)\sqrt{x}\,dx$$
$$\qquad\qquad=-\frac{5}{2}\int\_0^\infty x\frac{d}{dx}[\psi'(x)]^2\,dx=\frac{5}{2}\int\_0^\infty [\psi'(x)]^2\,dx.$$
On the other hand,
$$\int\_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}}\,dx=\int\_0^\infty \psi(x)\psi''(x)\,dx=-\psi'(0)-\int\_0^\infty[\psi'(x)]^2\,dx.$$
Hence
$$\int\_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}}\,dx=-\frac{5}{7}\psi'(0).$$




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$^\ast$ See, for, example, page 17 of [Roi Baer's notes](http://vintage.fh.huji.ac.il/~roib/LectureNotes/DFT/dft02.pdf)


