Question

Compute
\[\int_0^1 \frac{x^r-1}{\ln(x)} \, dx
\]
Where \(r>0\) is a real number.

Answer 1

Level Calculus III

Answer 2

I think DUI/feynman trick will do here \[\begin{align}
I(r)&=\int_0^1 \frac{x^r-1}{\ln(x)} \, dx\\~\\
I'(r)&=\int_0^1 x^r \, dx = \dfrac{1}{r+1}\\~\\
I(r) &= \ln(r+1) + C
\end{align}\]

\(I(0) = 0 \implies C = 0\)

Answer 3

I dare someone to try to find a method that doesn't involve differentiation under the integral sign. =P

Answer 4

(I am not claiming that another method exists, but I only suspect there might be, just for fun)

Answer 5

Well I am really technically doing the same thing as ganeshie I believe, but I'll go ahead and post it anyways. Originally I was looking to try to do some thing with the geometric series but since r is any real number that really won't work. At any rate, here we go: \[\Large x^r-1 = x^r-x^0 = x^y |_0^r = \int\limits_0^rx^y \ln x dy\] Replace this in the integral as \[\Large \int\limits_0^1 \int\limits_0^r x^y dydx = \int\limits_0^r \int\limits_0^1 x^ydxdy \\ \Large \int\limits_0^r \frac{1}{y+1}dy\] Yeah, just sort of ended up looking at it inside out oh well lol.

Answer 6

By a change of variable, this can be transformed to integral that I posted in my previous problem.

Answer 7

\[
x= e^u\\
dx= e^u du \\
\int_0^1 \frac{x^r-1}{\ln(x)} \, dx= \int_0^\infty \frac{e^{-(r+1)u}- e^{-u} }{u} \, du =\ln(r+1)
\]

Answer 8

Ahh I see this problem is a nice application of frullani's thm
http://math.stackexchange.com/questions/61828/proof-of-frullanis-theorem