Let `f(x)` be a continuous function `AAx in R ,` except at `x=0,` such that `int_0^a f(x)dx`, `ain R^+` exists. If `g(x)=int_x^a(f(t))/t dt `, prove that `int_0^af(x)dx=int_0^ag(x)dx`

Explanation

We have `g(x)=int_(x)^(a)(f(t))/t dt` ………………..1
Differentiating both sides w.r.t `x`, we get
`g'(x)=-(f(x))/x` or `f(x)=-xg'(x)`
or `int_(0)^(a)f(x)dx=-int_(0)^(a)dg(x)dx`
`=-x g(x)|_(0)^(a)+int_(0)^(a)g(x)dx`
`=-ag(a)+int_(0)^(a)g(x)dx`
`=int_(0)^(a)g(x)dx` [as form 1 `g(a)=0`]