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We are given parameters $a > 0, b > 0$.
Task is to integrate that: $\displaystyle \int_0^1 {\frac {x^a-x^b} {\ln x} dx}$.
I have tried approaching problem from different angles with no luck. I tried integration by parts(tried all combinations of possible $v$ and $u$), u-substitution with no luck.
Also I tried to integrate this two similar terms separately.

Tried to get some idea of how to go from answer, got nice answer from MATLAB: $\displaystyle \ln{\frac{a+1}{b+1}}$, but no idea how to reach it.

I would appreciate some suggestions.

Harry Peter
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Dmitri K
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  • For fixed $b$, consider $$I(a) = \int_0^1 \frac{x^a-x^b}{\ln x},dx.$$ Differentiate, $I'(a) = \dotsc$. Note $I(b) = 0$. – Daniel Fischer May 04 '14 at 16:33
  • @Norbert Generally, integral the same but limits and parameter assumptions are different. So I would not consider it as exact duplicate. Thank you for the links and suggestion, though. I got to sit on it. – Dmitri K May 04 '14 at 16:42
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    Substitute $t = x^{-1}$ and use the method in the comment by @Norbert. – Hans Engler May 04 '14 at 16:45
  • @Venus I think this question is a generalization of the other. It's worth linking them but I think they both seem to have distinct contributions (looking at the answers each got). – David K Dec 20 '14 at 06:18
  • @DavidK IMO, they're both essentially the same. Let the community decide it whether this is a duplicate of the another or not – Venus Dec 20 '14 at 06:53

3 Answers3

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First notice that: $$I = \int_{0}^{1} \frac{x^b-x^a}{\ln x} dx = \int_{0}^{1} \Big[\int_{a}^{b} x^y dy\Big]dx $$ The function $f(x,y)=x^y$ is continous in the set $[0,1]\times[a,b]$, therefore:

$$ I= \int_{0}^{1} \Big[\int_{a}^{b} x^y dy\Big]dx = \int_{a}^{b} \Big[\int_{0}^{1} x^y dx\Big]dy = \int_{a}^{b} \frac{1}{y+1}dy = \ln \Big(\frac{b+1}{a+1} \Big)$$

Hamdi Dervodeli
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If we let $t=-\ln x$ then, $dx =-e^{-t}dt$ and the result follows frullani's theorem $$ \int_{0}^{1} \frac{x^b-x^a}{\ln x} dx =- \int_{0}^{\infty} \frac{e^{-tb}-e^{-ta}}{t} e^tdt=-\int_{0}^{\infty} \frac{e^{-t(b+1)}-e^{-t(a+1)}}{t} dt= \ln\left(\frac{b+1}{a+1}\right)$$

Here is the Frullani's theorem :Proof of Frullani's theorem

Also see this https://math.stackexchange.com/q/2521520

Guy Fsone
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First of all I got to say that I like user140835's solution much more. It has this strict pure beauty of mathematical derivations.
But, from one hand, I have a feeling that I would fail to recognize applicability of that approach on similar problems. From other hand, I wanted to solve it by myself with another approach suggested in comments.

So I will leave my solution here, just in case if it will be helpful for somebody.

Performing u-substitution: $u = \frac 1 x$. Hence $dx = \frac {-du} {u^2}$
$\displaystyle -\int \frac {u^{-a}-u^{-b}}{u^2 ln{\frac 1 u}}du = I\space \Rightarrow \space \frac {dI(a)} {da} = \frac {u^{-a} \cdot ln(u)} {u^2ln{\frac 1 u}}$

Now, integrate received expression over u: $\int_\infty^0 \frac{dI(a)}{da}du=\int_\infty^0 \frac {u^{-a} \cdot ln(u)} {u^2ln{\frac 1 u}}du=-\frac {u^{-(a+1)}} {a+1} |_\infty^0=\frac {-1} {a+1}$.
Now we have to find antiderivative of that, which is $I(a)= -ln(a+1)$.
Following analogous steps for $I(b)$ we get the answer: $log(\frac{b+1}{a+1})$

Dmitri K
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