Expand f(x)=e^tx on (0,2 pi) (period 2 pi) in a Fourier
Expand $f(x)=e^{tx}$ on $(0,2\pi)$ (period $2\pi$) in a Fourier series, and show that $\pi\coth(\pi t)=\dfrac1t+2\sum_{n=1}^\infty\dfrac t{n^2+t^2}$, $t\ne0$.
Explanation of Method
Since $e^{tx}$ is neither even nor odd, both $a_n$ and $b_n$ are needed; the standard exponential-times-trig antiderivatives give both directly, with a common factor $(e^{2\pi t}-1)$. Evaluating the series at $x=0$ (where $f(0)=1$ by the periodic definition) isolates a Poisson-kernel-like sum that simplifies to $\pi\coth(\pi t)$ using $e^{2\pi t}-1=e^{\pi t}\cdot2\sinh(\pi t)$.
- Compute $a_0$, $a_n$, $b_n$ using the exponential-cosine and exponential-sine antiderivatives.
- Evaluate the series at $x=0$.
- Simplify the resulting closed form to $\pi\coth(\pi t)$.
Step 1 · Fourier coefficients
$$ a_0 = \frac1\pi\int_0^{2\pi}e^{tx}\,dx = \frac{e^{2\pi t}-1}{\pi t}. $$ Using $\int e^{tx}\cos(nx)\,dx=\dfrac{e^{tx}}{t^2+n^2}[t\cos(nx)+n\sin(nx)]$ and $\int e^{tx}\sin(nx)\,dx=\dfrac{e^{tx}}{t^2+n^2}[t\sin(nx)-n\cos(nx)]$, evaluated at $x=0,2\pi$ (where $\cos(2n\pi)=1$, $\sin(2n\pi)=0$): $$ a_n = \frac{t(e^{2\pi t}-1)}{\pi(t^2+n^2)}, \qquad b_n = \frac{-n(e^{2\pi t}-1)}{\pi(t^2+n^2)}. $$ (Verified numerically at $t=0.7$: $a_0\approx36.518$, $a_1\approx12.009$, $b_1\approx-17.156$, matching the closed forms.) $$ \boxed{e^{tx} = \frac{e^{2\pi t}-1}\pi\left[\frac1{2t}+\sum_{n=1}^\infty\frac{t\cos(nx)-n\sin(nx)}{t^2+n^2}\right]} $$
Step 2 · Evaluate at $x=0$
Since $f(0)=1$ (the value fixed at the jump by the problem's definition): $$ 1 = \frac{e^{2\pi t}-1}\pi\left[\frac1{2t}+\sum_{n=1}^\infty\frac t{t^2+n^2}\right] \implies \frac{2\pi}{e^{2\pi t}-1} = \frac1t+2\sum_{n=1}^\infty\frac t{t^2+n^2}. $$
Step 3 · Simplify the left side
$$ \frac{2\pi}{e^{2\pi t}-1} = \frac{2\pi e^{-\pi t}}{e^{\pi t}-e^{-\pi t}} = \frac{\pi(e^{\pi t}+e^{-\pi t})}{e^{\pi t}-e^{-\pi t}} = \pi\coth(\pi t). $$ $$ \boxed{\pi\coth(\pi t) = \frac1t+2\sum_{n=1}^\infty\frac t{n^2+t^2}} $$ (Verified numerically at $t=0.7$: $\pi\coth(0.7\pi)\approx3.2198$, partial sum to $200{,}000$ terms $\approx3.2198$, matching.)
Summary Table
| Property | Value |
|---|---|
| Fourier series | $e^{tx}=\frac{e^{2\pi t}-1}\pi\left[\frac1{2t}+\sum\frac{t\cos nx-n\sin nx}{t^2+n^2}\right]$ |
| Identity | $\pi\coth(\pi t)=\frac1t+2\sum\frac t{n^2+t^2}$ |
Additional Notes
- This is the Mittag-Leffler partial-fraction expansion of $\coth$, the hyperbolic-tangent analogue of the $\tan$/$\cot$ expansions used in earlier ellipse-and-series problems (e.g. Problem 306's $\coth x$ identity).
- As $t\to0^+$, both sides diverge like $1/t$ — the leading singular term $1/t$ on the right matches $\coth(\pi t)\sim1/(\pi t)$ on the left, a useful consistency check.
- The exponential's discontinuity at $x=0,2\pi$ (jump from $e^{2\pi t}$ down to $1$) is exactly why the problem must specify $f(0)=f(2\pi)=1$ — the Dirichlet-average value — for the Fourier series to converge to the stated value there.