Integral identity?

Question

How do you prove this integral identity?

∫{(x^a-1)/log(x)}dx=log(a+1) where the integral is from 0 to 1

Maple doesn't even give me the answer, so (a) a calculator will not work, and (b) it can not be proven as easily as using integration by parts.

Even if Maple (or a calculator) gave me the answer, that would not be proving it.

well, since (x^(a-1)) doesn't work, it kinda has to be (x^a-1). Basically, go with the one that works :).

How about trying to show it is true for a=1 first. I think that is a good start.

Scott, you did it in one way, but not the way that I think is the "best" (of course this is subjective). I don't really "like" your method because Li and Ei are some strange functions and this can be done with no knowledge of such functions (or their attributes). A suggestion: let the integral equal phi(a) and differentiate phi(a) with respect to a. What happens?

Pessimist_Athiest it is an identity, and I will try and give clues until someone can figure it out.

Scott, this was a homework problem for me about four years ago, and the reason I bring it up is because it is a method of integration that I find amazing. Sadly it is very difficult to find situations to use this method smartly (this is the only problem I know of).

I will eventually post my proof when I choose a best answer: I hope you guys enjoy it as much as I did/do.

And I just checked it (so that I can say I did, cause I know that it is true :)) on Maple and all the values from 0 to 10 work. Again it's ∫[((x^a)-1)/ln(x)]dx=ln(a+1)

Answer 1

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.

Answer 2

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?

Answer 3

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.

Answer 4

Try integration by parts.

Answer 5

use the calculator?
easy way out: substitue for "a"