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For period $2$, use the Fourier series conventionThe Fourier coefficients have prefactor $1$ on $[-1,1]$. First, $a_0={\int }_{-1}^1{e}^{x}dx=2\sinh 1$. For $q=n\pi$ the useful antiderivatives areAt the endpoints $\sin (n\pi )=0$ and $\cos (n\pi )=(-1{)}^n$, soConsequently, on $-1<x<1$,The Dirichlet-Jordan convergence theorem justifies equality at every point of continuity of the periodic extension. At odd integers the extension has a jump discontinuity between $e$ and $e^{-1}$, so the series converges to their average $\cosh 1$, regardless of any assigned endpoint value. This is the $\kappa =L=1$ case of the Fourier series of a periodically extended exponential.

Assume $L>0$. For any nonzero integer $k$,For $m\ne n$, the product-to-sum identity expresses the integrand as one half the sum of cosines with nonzero frequencies $m-n$ and $m+n$, so the integral is zero. For $m=n>0$, use ${\cos }^{2}u=(1+\cos 2u)/2$ to get $L$. Finally, for $m=n=0$ the integrand is one. The complete Fourier orthogonality formula isThe constant mode has twice the squared norm of a positive-frequency cosine mode.

At the origin the Hessian matrix isThus the second-derivative test gives a saddle point. Its opposite signs can also be checked without a test formula: $g(x,0)=x^2{e}^{-x^2/2}>0$ for $x\ne 0$, whereas $g(0,y)=-y^2{e}^{-y^2/2}<0$ for $y\ne 0$. Both types of values occur arbitrarily close to the zero value at the origin, excluding either a local maximum or a local minimum.

The product rule gives the two partial derivativesThey both vanish at the origin. The exponential function is never zero, so any further stationary point must satisfy $x=0$ or $x^2-y^2=2$, and $y=0$ or $y^2-x^2=2$. Having both coordinates nonzero would require both opposite differences to be $2$, which is impossible. If $y=0$, the nonzero choices are $x=\pm \sqrt{2}$; if $x=0$, they are $y=\pm \sqrt{2}$. Hence the full list is