The trial is a product of two identical 1s orbitals, one per electron. Its energy splits into one-electron pieces (kinetic + nuclear attraction, doubled) plus the two-electron repulsion. Every piece is an average over $|\varphi {|}^2$, so write each as an explicit integral.

Normalization. The factor $({\zeta }^3/\pi {)}^{1/2}$ is fixed by demanding $⟨\varphi |\varphi ⟩=\int |\varphi (\mathbf{r}){|}^2{d}^{3}r=1$:

The one identity doing all the work below is ${\int }_0^{\infty }{r}^n{e}^{-ar}dr=n!/a^{n+1}$.

Nuclear attraction, per electron — the operator $-Z/r$ averaged over $|\varphi {|}^2$:

With $Z=2$ and two electrons: $⟨V_{ne}⟩=2(-Z\zeta )=-2Z\zeta =-4\zeta$.

Kinetic energy. Use the gradient form $⟨T⟩=\frac{1}{2}\int |\nabla \varphi {|}^2{d}^{3}r$ — there is the absolute value, $|\nabla \varphi {|}^2$, not $\nabla \varphi$. Since $\nabla \varphi =-\zeta \hat{\mathbf{r}}\varphi$, we get $|\nabla \varphi {|}^2={\zeta }^2|\varphi {|}^2$:

Two electrons: $⟨T⟩=2\cdot \frac{1}{2}{\zeta }^2={\zeta }^2$.

Electron–electron repulsion — the one genuinely two-electron integral, over $|\varphi ({\mathbf{r}}_1){|}^2|\varphi ({\mathbf{r}}_2){|}^2$. Expanding $1/r_{12}$ in Legendre polynomials and integrating gives the standard 1s result:

Assemble the three pieces:

Predict before reading on. You have $E(\zeta )$. Before minimizing: each electron screens the nucleus from the other, so will the optimal exponent come out above or below the bare charge $Z=2$?

Check. $-2.8477>-2.9037$ ✓ — above exact, as the variational principle demands. And ${\zeta }^{\star }=1.6875<Z=2$: each electron sees a screened nucleus, with screening constant $\sigma =Z-{\zeta }^{\star }=0.3125$ — essentially Slater’s empirical $0.30$.