Prove that $\int_0^1\frac{x^y-1}{\log x}\mathrm dx=\log(1+y)$ The title says it all - I currently can't find a good way to start. Tried rewriting it into a line integral, but I really don't see a way to solve this right now. I'd appreciate any hints.
 A: Let $$I = \int_{0}^{1}\frac{x^y-1}{\ln x}dx = \int_{0}^{1}\int_{0}^{y}x^{z}dzdx$$
So $$I = \int_{0}^{y}\int_{0}^{1}x^{z}dxdz = \int_{0}^{y}\left[\frac{x^{z+1}}{z+1}\right]_{0}^{1}dz = \int_{0}^{y}\frac{1}{z+1}dz$$
So $$I = \left[\ln|z+1|\right]_{0}^{y} = \ln|y+1|$$
A: There's quite a few ways to prove this, but I personally like this one.
$$I(y)=\int_0^1\frac{x^y-1}{\log x}\mathrm dx$$
$$\frac{\mathrm dI(y)}{dy}=\frac{\mathrm d}{\mathrm dy}\int_0^1\frac{x^y-1}{\log x}\mathrm dx=\int_0^1\frac{\partial}{\partial y}\left(\frac{x^y-1}{\log x}\right)\mathrm dx=\int_0^1x^y\mathrm dx=\frac{1}{1+y}$$
$$I(y)=\int\frac{\mathrm dy}{1+y}=\log(1+y)+C$$
Seeing that $I(0)=0=C$ gives the desired result.
A: $$\int_0^1\frac{x^y-1}{\log x}\mathrm dx=\int_0^1 \frac{x^y}{\log x}dx - \int_0^1 \frac{1}{\log x}dx = li (x^{y+1}) -li(x)|^1_0$$
$li(x)$ series expansion at $x = 1$:
$$li (x) = log(x-1) + \gamma + O(x-1) $$
$$li(0) = 0$$
$li(1)$ is undefined
the upper limit should be 
$$lim_{x \to 1} (li(x^{y+1}) -li(x)) = lim_{x \to 1}  [log(x^{y+1}-1) - log(x-1) + O(x-1)]  =lim_{x \to 1} [log(\frac{x^{y+1} - 1}{x-1}) +O(x-1)] $$
By L-Hospital rule: 
$$lim_{x \to 1} \frac{x^{y+1} - 1}{x-1} = (y+1)x^y$$
In conclusion:
$lim_{x \to 1} (li(x^{y+1}) -li(x)) = lim_{x \to 1} [log((y+1)x^y) +O(x-1)] 
 = log(y+1)$
$$\int_0^1\frac{x^y-1}{\log x}dx =log(y+1) $$
A: A simple way is to enforce the substitution $x=e^{-t}$ then apply Frullani's theorem.
