A Fourier Series

We find the Fourier series for y = e ^ ( 2 `pi k).

The coefficients ak are found from the formula for the integral of e^(ax) cos(bx), as indicated below.
When k is odd, the argument k * `pi of the sine and cosine functions will lie on the negative x axis. The value of the cosine will therefore be -1, and we will obtain the first of the two results below.
When k is even, the argument k * `pi of the sine and cosine functions will lie on the positive x axis, giving us the second of the results below.

Thus the coefficients of e^(2 `pi^2) and e^(-2 `pi^2) will be of opposite sign, their signs will alternate with even and odd k.

A similar calculation for bk results in an analogous result.

Note that in this case the sine term has factor 2 `pi, while the cosine term has factor k.
The sine term is 0 at the limits, so we do not end up with the factor 2 `pi in our coefficient. We rather end up with the factor k.

We easily evaluate a0.

We obtain the series indicated below.

The coefficient of cos(kx) is 4 sinh(2 `pi^2) * [ (-1)^k / (4 `pi^2 + k^2) ], while that of sin(kx) is -2 k / `pi sinh(2 `pi^2) * [ (-1)^k / (4 `pi^2 + k^2) ].

The (-1^k) / ( 4 `pi^2 + k^2 ) should perhaps be factored out of the two summations.

We note that we could then obtain kth term 4 cos(kx) - 2 k / `pi sin(kx), of the form A cos(kx) + B sin(kx), which could be simplified using a simple trigonometric identity into the form sin(kx + `phi) for an appropriate `phi.