Question

Evaluate the following integrals : `int_0^1 log(1/x - 1)*dx`

Answer 1

Let I = `int_0^1 log(1/x - 1)*dx`

∴ I = `int_0^1 log((1 - x)/x)*dx`       ...(i)

= `int_0^1 log[(1 - (1 - x))/(1 - x)]*dx     ...[because int_0^"a" f(x)*dx = int_0^"a" f("a" - x)*dx]`

I = `int_0^"a" log(x/(1 - x))*dx`    ...(ii)

Adding (i) and (ii), we get

2I = `int_0^1 log((1 - x)/x)*dx + int_0^1 log(x/(1 - x))*dx`

= `int_0^1[log  ((1 - x)/x) + log (x/(1 - x))]*dx`

= `int_0^1 log ((1 - x)/x  xx x/(1 - x))*dx`

= `int_0^1 log 1*dx`

∴ 2I = `int_0^1 0*dx`
∴ I = 0.

Answer 2

Let I = `int_0^1 log(1/x - 1)*dx`

= `int_0^1 log((1 - x)/x)*dx`

= `int_0^1 [log(1 - x) - logx]*dx`             ...(1)

We use the property `int_0^a f(x)*dx = int_0^a f(a - x)*dx`

Here, a = 1
Hence in I, changing x to 1 – x, we get

I = `int_0^1 [log |1 - (1 - x)| - log(1 - x)]*dx`

= `int_0^1 [logx - log(1 - x)]*dx`

= `-int_0^1 [log (1 - x) - logx]*dx`

= – 1            ...[By (1)]
∴ 2I = 0
∴ I = 0.