integration of log(1/2 sin2x)

Question

definite integration of log(1/2 sin2x)

Answer 1

Careful inspection of this integral reveals that log(½sin(2x)) has no meaning when sin(2x)<0, because the log of a negative number is not possible. This implies that the indefinite integral cannot be evaluated because of this anomaly. However, log|½sin(2x)| is computable. Graphically, the integral represents the area between the curve and the x-axis. The antiderivative is usually what is meant by the integral, and I don't know how to find the function whose derivative is log(½sin(2x)). But I can give a numerical result of the integral of log|½sin(2x)| by considering the definite integral in the interval [0,π/2]. This is the area (call it A) between the curve and the x-axis over this interval. The interval [π/2,π] is, by symmetry, an equal area A, so the area in [0,π]=2A.

A=∫₀^π/2log(½sin(2x))dx. I assume that log is the same as ln (natural log).

A=∫₀^π/2ln(sin(2x))dx-∫₀^π/2ln(2)dx=∫₀^π/2ln(sin(2x))dx-ln(2)[x]₀^π/2=∫₀^π/2ln(sin(2x))dx-½πln(2)

Let u=2x, then du=2dx, dx=du/2; when x=0, u=0 and when x=π/2, u=π.

Let B=∫₀^π/2ln(sin(2x))dx=½∫₀^πln(sin(u))du. But from symmetry we know that ∫₀^πln(sin(u))du=2∫₀^π/2ln(sin(u))du. This can also be expressed as 2∫₀^π/2ln(sin(x))dx.

So B=½(2∫₀^π/2ln(sin(x))dx)=∫₀^π/2ln(sin(x))dx). So A=B-½πln(2).

A can also be expressed:

A=∫₀^π/2ln(sin(x)cos(x))dx, because sin(2x)=sin(x)cos(x).

A=∫₀^π/2ln(sin(x))dx+∫₀^π/2ln(cos(x))dx=B+∫₀^π/2ln(sin(π/2+x))dx, because sin(π/2+x)=cos(x).

Let v=π/2+x, dv=dx, and when x=0, v=π/2 and when x=π/2, v=π.

∫₀^π/2ln(cos(x))dx=∫_π/2^πln(sin(v))dv, which, because of symmetry, is the same area as ∫₀^π/2ln(sin(x))dx=B.

Therefore A=2B=B-½πln(2). B=-½πln(2). This area is below the x-axis, which explains why it's negative.

A=2B=-πln(2). The actual value of the area beneath the x-axis is πln(2). If we extend this to the interval [0,π] we get 2πln(2).

The indefinite integral would therefore be 2πnln(2)=4.3552n approx. You could think of n as the integration constant and the interval over which the definite integral is applied would be [0,nπ]. More generally, for the interval [-mπ,nπ] integral (area)=2π(m+n)ln(2) beneath the x-axis (so actually -2π(m+n)ln(2)). This is the indefinite integral of log|½sin(2x)|.