Question

Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`

Answer 1

Let I = `int_0^1 logx/sqrt(1 - x^2)dx`

Put x = sin θ,

∴ dx = cos θ dθ

When x = 0, θ = 0 and

When x = 1, θ = `π/2`

∴ I = `int_0^(π/2) (logsinθ)/sqrt(1 - sin^2θ) * cosθ  dθ`

= `int_0^(π/2) (log sinθ)/cosθ * cosθ  dθ`

I = `int log sin θ  θ`   ...(1)

= `int_0^(π/4) [log sin θ + log sin(π/2 - θ)]dθ`   ...`{∵ int_0^(2a) f(x)dx = int_0^a[f(x) + f(2a - x)]dx}`

I = `int_0^(pi/4) [log sin θ + log cos θ]dθ`

= `int_0^(pi/4) log(sin θ cos θ)dθ`

= `int_0^(pi/4) log((sin 2θ)/2)dθ`

= `int_0^(pi/4) log sin 2θ dθ - int_0^(pi/4) log 2 dθ`

In first integral,

Put 2θ = t,

∴ dθ = `(dt)/2`

When θ = 0, t = 0

and When θ = `π/4`, t = `π/2`

∴ I = `int_0^(π/2) log sin t * dt/2 - [θ log 2]_0^(pi/4)`

∴ I = `1/2 int_0^(π/2) log sin t  dt - [π/4 log 2]`

∴ I = `1/2 int_0^(π/2) log sin θ  dθ - π/4 log 2`   ...(Definite integral is independent of variables.)

∴ I = `1/2 I - π/4 log 2`  ...[From (1)]

∴ `1/2 I = -π/4 log 2`

∴ I = `-π/2 log 2`

= `π/2 log (1/2)`

Hence proved.