Evaluating $ \int_0^1 \frac{x^a - 1}{\ln{}x} dx$

Question

I know how to evaluate using Leibnitz Rule

I was trying to see if I could solve it without that

$\displaystyle \int_0^1 \frac{x^a - 1}{\ln{}x} dx$ = $\displaystyle \int_0^1 \frac{x^a}{\ln{}x} dx$ - $\displaystyle \int_0^1 \frac{1}{\ln{}x} dx$

$\displaystyle \int_0^1 \frac{x^a}{\ln{}x} dx$ - $\displaystyle \int_0^1 \frac{1}{\ln{}x} dx$ = $\displaystyle \int_0^1 \frac{(a+1)x^a}{\ln{}x^{a+1}} dx$ - $\displaystyle \int_0^1 \frac{1}{\ln{}x} dx$

Substituting $x^{a+1} = t$ in the 1st integral we get $\displaystyle \int_0^1 \frac{1}{\ln{}t} dt$ - $\displaystyle \int_0^1 \frac{1}{\ln{}x} dx$ = $0$

Where am I going wrong??

Also any other ways to evaluate this?

Answer 1

Indeed, the issue you're running into with your approach is that you have taken a convergent integral and split it up into two divergent integrals.

For another approach we first define $$ I=\int_0^1\frac{x^a-1}{\log x}\,\mathrm dx. $$ Expanding $x^a=e^{a\log x}$ yields $$ I=\sum_{k=1}^\infty\frac{a^k}{k!}\int_0^1\log^{k-1}x\,\mathrm dx=-\sum_{k=1}^\infty\frac{(-a)^k}{k!}\int_0^1(-\log x)^{k-1}\,\mathrm dx. $$ A substitution of $u=-\log x$ turns the remaining integral into a gamma function giving $$ I=-\sum_{k=1}^\infty\frac{(-a)^k}{k!}(k-1)!=-\sum_{k=1}^\infty\frac{(-a)^k}{k}=\log(1+a), $$ which follows form the series expansion of the logarithm.

Answer 2

Note that, if $I(a)$ is the integral we're looking for, $$\frac{dI}{da}(a)=\int_0^1x^a dx = \frac 1 {1+a}$$ Thus $$I(a)-I(0) = I(a)=\log(1+a)$$