Question

Show that `int_0^1(x^a-1)/logx dx=log(a+1)`

Answer 1

let I = `int_0^1(x^a-1)/logx dx`
Taking ‘a’ as parameter ,

I(a) = `int_0^1(x^a-1)/logx dx` -------- (1)

differentiate w.r.t a ,

`(dI(a))/(da)=d/(da)int_0^1(x^a-1)/logx dx`

`therefore (dI(a))/(da)=int_0^1del/(dela)(x^a-1)/logx dx`  ………{ D.U.I.S f(x)}

`therefore (dI(a))/(da)=int_0^1(x^a.logx)/logx dx`     .......... `{(dx^a/da)=x^a.loga}`

`therefore (dI(a))/(da)=int_0^1x^a dx`

`therefore (dI(a))/(da)=[x^(a+1)/(a+1)]_0^1`

`therefore (dI(a))/(da)=1/(a+1)-0`

`therefore (dI(a))/(da)=1/(a+1)`

now , integrate w.r.t a,

`"I(a)"=int  1/(a+1)da`

`"I(a)"=log (a+1)+c` -------- (2)

where c is constant of integration
put a = 0 in eqn (1),

`"I"(0)=int_0^1 0 dx=0`

And
From eqn (2), I(0)= c
∴ c = 0
∴ I = log(a+1) 

Hence proved.