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Show ∫ x log(sin x) dx

Short answer

Use symmetry to get 2I = π∫_0^π log(sin x)dx = 2π∫_0^{π/2}log(sin x)dx. Since ∫_0^{π/2}log(sin x)dx = -(π/2)log 2, we obtain I = (π²/2)log(1/2).

Show that ∫₀^π x log(sin x) dx = (π²/2) log(1/2).

Let I = ∫₀^π x log(sin x) dx. Using ∫₀^a f(x) dx = ∫₀^a f(a − x) dx:

I = ∫₀^π (π − x) log(sin x) dx [since sin(π − x) = sin x]

Add the two forms:

2I = π ∫₀^π log(sin x) dx = 2π ∫₀^(π/2) log(sin x) dx [sin(π − x) = sin x, symmetric about π/2]

So I = π J, where J = ∫₀^(π/2) log sin x dx.

Find J. J = ∫₀^(π/2) log cos x dx as well (replace x by π/2 − x). Add:

2J = ∫₀^(π/2) log(sin x cos x) dx
   = ∫₀^(π/2) [log sin 2x − log 2] dx
   = ∫₀^(π/2) log sin 2x dx − (π/2) log 2

Put 2x = u: ∫₀^(π/2) log sin 2x dx = (1/2)∫₀^π log sin u du = ∫₀^(π/2) log sin u du = J. So 2J = J − (π/2) log 2, hence

J = −(π/2) log 2.

I = πJ = −(π²/2) log 2

I = (π²/2) log(1/2). Shown.

✓ Answer checked by Claude Opus 5.5 (high effort)

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