Integral identity?

Let F(a) be the value of that integral. Take the derivative with respect to a to get

F'(a)=int x^a * log(x)/log(x) dx

=int x^a dx

=1/(a+1) [1^(a+1) -0^(a+1)]

=1/(a+1).

Now take the anti-derivative to get

F(a)=log(a+1) +C.

But, clearly if a=0, we are integrating 0, so C=0.

This gives the result.

The only point at issue is whether differentiating under the integral sign is legitimate. In spite of appearances, the function inside the original integral is continuous (or can be extended to be continuous) and clearly the partial derivative is continuous in both x and a, so the result follows.

Hey, is or was this one of YOUR homework assignments?

You're a good one for symantics of the questions too, and as you didn't ask to prove it, but how to prove it, here is AN answer, and perhaps a hint for others.

Notice that

∫ (x^a -1)/log(x) dx = ∫(x^a)/log(x) dx - ∫(1/log(x)) dx

let y=log(x) → x = e^y , dx = e^y dy

then ∫ = ∫ (e^(ay))/y * e^y dy - ∫(1/log(x)) dx

= ∫ (e^[(a+1)y])/y dy - ∫(1/log(x)) dx

= Ei[(a+1)y] - Li[x]

=Ei[(a+1)log(x)] - Li[x]

use Li[z]=Ei[log(z)],

and go from there...

Eulercrosser, now go check my solution to:

Why is the diff of 2 numbers made by a different combinations of the same digits, equal to a multiple of 9?

Yeah, I don't think this is an identity. If it is, it sure as hell can't be written right. I checked it for a={0,1...10} and didn't get a single truth value. What gives?

I think you were implying a proof by mathematical induction, but if it fails for S(0), then you can't make any kind of proof.

Try integration by parts.

use the calculator?

easy way out: substitue for "a"