Expand the periodic function f(x)=ln |sin dfrac x2 |
Expand the periodic function $f(x)=\ln\left|\sin\dfrac x2\right|$ in a Fourier series.
Explanation of Method
$f$ is even (since $|\sin(-x/2)|=|\sin(x/2)|$), so only cosine terms appear. Both $a_0$ and $a_n$ reduce, via the substitution $u=x/2$, to the classical integrals $\int_0^{\pi/2}\ln(\sin u)\,du=-\frac\pi2\ln2$ and $\int_0^{\pi/2}\ln(\sin u)\cos(2nu)\,du=-\frac\pi{4n}$. The result also follows directly from the real part of the standard series $\ln|1-e^{ix}|=-\sum\cos(nx)/n+\ln2$-type identity.
- Note evenness, so $b_n=0$.
- Compute $a_0$ via the classical log-sine integral.
- Compute $a_n$ via the companion integral, and cross-check via the complex-logarithm identity.
Step 1 · Symmetry and $a_0$
$f$ is even, so $b_n=0$. With $u=x/2$: $$ a_0 = \frac4\pi\int_0^{\pi/2}\ln(\sin u)\,du = \frac4\pi\left(-\frac\pi2\ln2\right) = -2\ln2. $$
Step 2 · Computing $a_n$
$$ a_n = \frac4\pi\int_0^{\pi/2}\ln(\sin u)\cos(2nu)\,du = \frac4\pi\left(-\frac\pi{4n}\right) = -\frac1n. $$ This matches the real part of the standard identity $\ln|1-e^{ix}|=\ln|2\sin\frac x2|=-\sum_{n=1}^\infty\frac{\cos(nx)}n$ (from $\operatorname{Re}\ln(1-e^{ix})=\operatorname{Re}\left[-\sum e^{inx}/n\right]$), which directly gives $\ln|\sin\frac x2|=-\ln2-\sum\cos(nx)/n$.
Step 3 · Fourier series and verification
$$ \boxed{\ln\left|\sin\frac x2\right| = -\ln2-\sum_{n=1}^\infty\frac{\cos(nx)}n} $$ (Verified numerically at $x=1.0,2.0$: partial sums to $300{,}000$ terms match the direct log value to 4+ decimal places. Independent check at $x=\pi/2$: LHS $=\ln(1/\sqrt2)=-\frac12\ln2$; RHS: $\cos(n\pi/2)$ vanishes for odd $n$, and for $n=2k$ gives $-\ln2-\sum(-1)^k/(2k)=-\ln2+\frac12\ln2=-\frac12\ln2$. ✓)
Summary Table
| Property | Value |
|---|---|
| $a_0$ | $-2\ln2$ |
| $a_n$ | $-1/n$ |
| Fourier series | $\ln|\sin\frac x2|=-\ln2-\sum\frac{\cos nx}n$ |
Additional Notes
- This series diverges logarithmically at $x=2\pi k$ (where $\sin\frac x2=0$), consistent with the $1/n$ (non-absolute) coefficient decay — the series is only conditionally convergent away from the singular points.
- The companion function $\ln|\tan\frac x2|$ (Problem 345) is built directly from this result and its shifted twin $\ln|\cos\frac x2|$, showing how logarithmic Fourier series combine algebraically just like their arguments.
- This is the real part of the classical complex-analysis identity $\ln(1-z)=-\sum z^n/n$ evaluated on the unit circle $z=e^{ix}$ — the same generating-function idea used throughout the Poisson-kernel problems (346–348).