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Is there a way to compute this integral, $$\int_0^1 \frac{x^k-1}{\ln x}dx =\ln({k+1})$$ without using the derivation under the integral sign nor transforming it to a double integral and then interchanging the order of integration. High school techniques only if possible.

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    $\begingroup$ Your integral expression above is wrong, simply try $k=0$. It should be $\int_0^1 \frac{x^k-1}{\ln x}dx =\ln(k+1)$ $\endgroup$
    – Pauly B
    Nov 20 '14 at 4:48
  • $\begingroup$ Thank you Pauly, I have corrected it. $\endgroup$ Nov 20 '14 at 4:50
  • $\begingroup$ I think there's a way to do this by writing a double integral. With one order of the integral signs, we get the LHS (left hand side) of your post; with another order, we get the RHS. Fubini then tells us that these must be equal. $\endgroup$ Nov 20 '14 at 5:03
  • $\begingroup$ @columbus8myhw, That is exactly what Idris refuses to accept. $\endgroup$ Nov 20 '14 at 5:05
  • $\begingroup$ Anything is a high school technique if you have the right attitude! Except for Iwasawa theory. Definitely not Iwasawa theory. $\endgroup$
    – MCT
    Nov 20 '14 at 5:15
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Here is a seemingly elementary solution: Using the following simple limit

$$ \ln x = \lim_{h\to 0} \frac{x^{h} - 1}{h} = \lim_{n\to\infty} \frac{x^{1/n} - 1}{1/n}, $$

it follows that

\begin{align*} \int_{0}^{1} \frac{x^{k} - 1}{\log x} \, dx &= \lim_{n\to\infty} \int_{0}^{1} \frac{x^{k} - 1}{n (x^{1/n} - 1)} \, dx \\ &= \lim_{n\to\infty} \int_{0}^{1} \frac{(y^{nk} - 1)y^{n-1}}{y - 1} \, dy, \quad (\text{substitute } x = y^{n}) \\ &= \lim_{n\to\infty} \int_{0}^{1} (1+y+\cdots +y^{nk-1})y^{n-1} \, dy. \end{align*}

Now using the following simple formula

$$ \int_{0}^{1} y^{i-1} \, dy = \frac{1}{i}, $$

we get the following identity

\begin{align*} \int_{0}^{1} \frac{x^{k} - 1}{\log x} \, dx &= \lim_{n\to\infty} \left( \frac{1}{n} + \cdots + \frac{1}{nk+n-1} \right) =\lim_{n\to\infty} \sum_{i=0}^{nk-1} \frac{1}{n+i}. \end{align*}

Modifying the summand a little bit, we have

\begin{align*} \int_{0}^{1} \frac{x^{k} - 1}{\log x} \, dx &= \lim_{n\to\infty} \sum_{i=0}^{nk-1} \frac{1}{n} \frac{1}{1+(i/n)} = \int_{0}^{k} \frac{dt}{1+t} = \ln(k+1). \end{align*}

P.s. All these steps are elementary except for one point. Can you figure out where it is?


EDIT. Elementary solution is not always the easiest solution as the essence is often shadowed by the technical intricacy involved. Nevertheless, here is a solution which uses only freshman-level calculus:

I will assume that $k$ is a positive integer (as I implicitly did in the first solution). Now let $n$ be another positive integer. Then by the substitution $x = y^{n}$ we get

$$ I := \int_{0}^{1} \frac{x^{k} - 1}{\log x} \, dx = \int_{0}^{1} \frac{(y^{nk} - 1)y^{n-1}}{\log y} \, dy. $$

For the solution, we claim the following:

Claim. For $0 < y < 1$ we have $y \leq \frac{y-1}{\log y} \leq 1$.

Proof. Let $f(y) = y \log y + 1 - y$ and $g(y) = y - 1 - \log y$. By differentiating once, we have $$ f'(y) = \log y \leq 0 \quad \text{and} \quad g'(y) = 1 - y^{-1} \leq 0.$$ This shows that both $f$ and $g$ are decreasing on $(0, 1]$. Since $f(1) = 0$ and $g(1) = 0$, we get both $$ f(y) \geq 0 \quad \text{and} \quad g(y) \geq 0 \quad \text{for} \quad 0 < y < 1. $$ The claim follows from these two inequalities. ////

Now write $I$ in the following form:

$$ I = \int_{0}^{1} \frac{(1 - y^{nk})y^{n-1}}{1 - y} \frac{y - 1}{\log y} \, dy. $$

Then by the claim we obtain

$$ \int_{0}^{1} \frac{(1 - y^{nk})y^{n}}{1 - y} \, dy \leq I \leq \int_{0}^{1} \frac{(1 - y^{nk})y^{n-1}}{1 - y} \, dy. $$

We know how to calculate both integrals and where they converge as $n \to \infty$ from the previous answer:

$$ \lim_{n\to\infty} \int_{0}^{1} \frac{(1 - y^{nk})y^{n}}{1 - y} \, dy = \lim_{n\to\infty} \int_{0}^{1} \frac{(1 - y^{nk})y^{n-1}}{1 - y} \, dy = \log(k+1). $$

Therefore the conclusion follows from the Squeezing Theorem.

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    $\begingroup$ Maybe interchanging the integral and limit. $\endgroup$ Nov 20 '14 at 5:37
  • $\begingroup$ @IdrisAddou Even my answer involves swapping the limit (of the infinite summation) and the integral. At the end of the day, all the "different" techniques relies on carefully and judiciously swapping two different limits. $\endgroup$
    – Adhvaitha
    Nov 20 '14 at 5:40
  • $\begingroup$ @sos440: You used this matter: $\int_{0}^{1}f(x)dx=\int_{0}^{1}\lim_{n\rightarrow \infty }f_{n}(x)dx=\lim_{n\rightarrow \infty }\int_{0}^{1}f_{n}(x)dx$ which is allowed under somme restriction as uniform convergence of the function sequence on the compact [0,1]. I think this is not a high school technique. Any way, i find it nice. Bay the way it is in essence the same way as that using double integral and then Fubini's thm. $\endgroup$ Nov 20 '14 at 5:44
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    $\begingroup$ @Idris That is exactly what I pointed out in my PS. In this case even uniform convergence fails, so we need the dominated convergence theorem. Thankfully, there is a way to circumvent this problem by using solely calculus techniques as I demonstrated in the edited version, but I do not think these solutions reveal the nature of our integral. $\endgroup$ Nov 20 '14 at 6:47
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    $\begingroup$ Correction: the integrand in my first solution does converge uniformly as $n \to \infty$. What I meant was that $1/(n(x^{1/n} - 1))$ does not converge uniformly as $n \to \infty$, but thanks to the factor $x^{k} - 1$ we indeed obtain the uniform convergence of $(x^{k} - 1)/(n(x^{1/n} - 1))$. $\endgroup$ Nov 20 '14 at 6:57
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$$ \begin{align} \int_0^1\frac{x^k-1}{\log(x)}\mathrm{d}x &=\int_0^\infty\frac{e^{-x}-e^{-(k+1)x}}{x}\mathrm{d}x\tag{1}\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-x}-e^{-(k+1)x}}{x}\mathrm{d}x\tag{2}\\ &=\lim_{\epsilon\to0^+}\left[\int_\epsilon^\infty\frac{e^{-x}}{x}\mathrm{d}x-\int_\epsilon^\infty\frac{e^{-(k+1)x}}{x}\mathrm{d}x\right]\tag{3}\\ &=\lim_{\epsilon\to0^+}\left[\int_\epsilon^\infty\frac{e^{-x}}{x}\mathrm{d}x-\int_{(k+1)\epsilon}^\infty\frac{e^{-x}}{x}\mathrm{d}x\right]\tag{4}\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^{(k+1)\epsilon}\frac{e^{-x}}{x}\mathrm{d}x\tag{5}\\ &=\lim_{\epsilon\to0^+}\int_1^{k+1}\frac{e^{-\epsilon x}}{x}\mathrm{d}x\tag{6}\\ &=\int_1^{k+1}\frac1x\mathrm{d}x-\lim_{\epsilon\to0^+}\int_1^{k+1}\frac{1-e^{-\epsilon x}}x\mathrm{d}x\tag{7}\\ &=\int_1^{k+1}\frac1x\mathrm{d}x\tag{8}\\[10pt] &=\log(k+1)\tag{9} \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto e^{-x}$
$(2)$: express as a limit so that we can separate the integrals
$(3)$: separate the integrals
$(4)$: substitute $x\mapsto x/(k+1)$
$(5)$: recombine integrals
$(6)$: substitute $x\mapsto\epsilon x$
$(7)$: separate the significant from the rest
$(8)$: Squeeze Theorem: $1-\epsilon x\le e^{-\large\epsilon x}\le1\implies0\le\int_1^{k+1}\frac{1-e^{-\large\epsilon x}}x\mathrm{d}x\le\epsilon k$
$(9)$: integrate

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Note that $x^k = e^{k \ln(x)}$. Hence, we have $$\dfrac{x^k-1}{\ln(x)} = \sum_{l=1}^{\infty} \dfrac{k^l \ln^{l-1}(x)}{l!}$$ Hence, our integral becomes $$\sum_{l=1}^{\infty} \dfrac{k^l}{l!} \int_0^1 \ln^{l-1}(x)dx = \sum_{l=1}^{\infty} \dfrac{k^l}{l!} \int_0^1 \ln^{l-1}(x)dx = \sum_{l=1}^{\infty} (-1)^{l-1}\dfrac{k^l}l = \ln(1+k)$$ since $$\displaystyle \int_0^1 \ln^{l-1}(x)dx = \int_{\infty}^0 (-t)^{l-1} e^{-t} (-dt)= (-1)^{l-1}\int_0^{\infty} t^{l-1}e^{-t} dt = (-1)^{l-1} (l-1)!$$


As pointed out in the comments, this is true for $-1<k\leq 1$. For $k> 1$, the proof can be easily extended. Set $$I(k) = \int_0^1 \dfrac{x^k-1}{\ln(x)}dx$$ We then have \begin{align} I(k) - I(k-1) & = \int_0^1 x^{k-1}\dfrac{(x-1)}{\ln(x)}dx = \sum_{l=1}^{\infty}\int_0^1 \dfrac{x^{k-1} \ln^{l-1}(x)}{l!}dx\\ & = \sum_{l=1}^{\infty}(-1)^l \dfrac{(1/k)^l}l = \ln(1+1/k) \text{ (since $1/k<1$)} \end{align} This now gives us $I(k) = \ln(1+k)$ for all $k$.

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  • $\begingroup$ Very nice calculation, thanks!!! $\endgroup$ Nov 20 '14 at 5:10
  • $\begingroup$ This derivation is only valid for $-1 < k < 1$, and perhaps also at $k = 1$ if you are careful, but the result itself is true for all $k > -1$. So this is not a complete explanation. To be fair, though, the original question was vague about the range of $k$ for which the equality was desired. $\endgroup$
    – KCd
    Nov 20 '14 at 5:11
  • $\begingroup$ @KCd Hopefully fixed now for all $k>-1$. $\endgroup$
    – Adhvaitha
    Nov 20 '14 at 5:26
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    $\begingroup$ Of course, swapping the infinite summation and the integral may be considered as non-high school stuff (depending on the high school you study). $\endgroup$
    – Adhvaitha
    Nov 20 '14 at 5:32
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I'm not expecting any upvote for this but surely this answer could also help someone!.

Let $$I(k)=\int_0^1{\frac{x^k-1}{\ln x}dx}$$

$$I'(k)=\int_0^1{\frac{(x)^k\ln x}{\ln x}dx}=\int_0^1{x^kdx}=\frac{x^{k+1}}{k+1}=\frac{1}{k+1}$$

$$I(k)=\ln(k+1)+C$$

Since we have $I(0)=0\implies \ln(1)+C=0\implies C=0$

$$I(k)=\int_0^1{\frac{x^k-1}{\ln x}dx}=\ln(k+1)$$

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    $\begingroup$ This is a great proof but not a very good answer. OP explicitly asks for a proof without this technique. $\endgroup$ Nov 20 '14 at 5:11
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    $\begingroup$ FYI, the "Leibniz Integral Rule" is precisely the method of differentiation under the integral sign, which was explicitly disallowed in the question. $\endgroup$
    – KCd
    Nov 20 '14 at 5:12
  • $\begingroup$ Hi there, thanks all of you for your interests in my question. I asked to do not use Leibniz rule nor Fubini's technic and I precise now to also do not use the change of variable $x=e^{-y}$ which convert the problem to the Frullani's integral. The raison is that all these ways to do are previously used in this site, so I asked new way! The way by Adhvaitha is new for me even if some restrictions are needed as KCd reported. Thanks for all. $\endgroup$ Nov 20 '14 at 5:20
  • $\begingroup$ @Integrator: IdrisAddou explained that the interest was in seeing a new way to solve the problem, where "new" means not previously known to the OP. I had already seen the power series method of Adhvaitha before, but evidently IdrisAddou had not, so it counts as new. $\endgroup$
    – KCd
    Nov 20 '14 at 5:35
  • $\begingroup$ @Integrator thanks for letting us know that what could be another and easy way to solve the integral in question. $\endgroup$
    – kaka
    Nov 20 '14 at 9:05
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Yet another answer. It is easy to see that $\displaystyle f(x)=\frac{x^k-1}{\log x}$, with $f(0)=0$ and $f(1)=k$ is a continuous fonction on $[0,1]$ for $k> 0$. Put now for $0<h<1$ $\displaystyle F(h)=\int_0^h f(t)dt$. We have $F(h)\to F(1)=I_k$ if $h\to 0$, $h<1$. But: $$F(h)=\int_0^{h}\frac{x^k}{\log x}dx-\int_0^h \frac{dx}{\log x}=\int_{h^{k+1}}^h \frac{-dx}{\log x}$$ By the change of variable $t^{k+1}=u$ in the first integral.

Now for $x\in [h^{k+1},h]$ we have $$h^{k+1}\frac{-1}{x\log x}\leq \frac{-x}{x\log x}\leq h\frac{-1}{x\log x}$$ where we have used that $\displaystyle \frac{-1}{x\log x}\geq 0$.

As $h^{k+1}\leq h$, integrating we get: $$h^{k+1}\int_{h^{k+1}}^h \frac{-1}{x\log x}dx\leq F(h)\leq h\int_{h^{k+1}}^h\frac{-1}{x\log x}dx$$ Hence $$h^{k+1}\log(k+1)\leq F(h)\leq h\log(k+1)$$ and if $h\to 1$, we get the result.

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  • $\begingroup$ Nice work! (+1) This solution is elementary yet loses no intuition. $\endgroup$ Nov 20 '14 at 14:51
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

$\ds{\int_{0}^{1}{x^{k} - 1 \over \ln\pars{x}}\,\dd x =\ln\pars{k + 1}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large\int_{0}^{1}{x^{k} - 1 \over \ln\pars{x}}\,\dd x} =-\int_{0}^{1}\pars{x^{k} - 1}\ \overbrace{\int_{0}^{\infty}x^{t}\,\dd t}^{\ds{=\ \color{#c00000}{-\,{1 \over \ln\pars{x}}}}}\ \,\dd x\ =\ -\int_{0}^{\infty}\int_{0}^{1}\pars{x^{k + t} - x^{t}}\,\dd x\,\dd t \\[5mm]&=\int_{0}^{\infty}\pars{{1 \over t + 1} - {1 \over t + k + 1}}\,\dd t =\left.\ln\pars{t + 1 \over t + k + 1}\right\vert_{\, t\ =\ 0}^{\, t\ \to\ \infty} =-\ln\pars{1 \over k + 1} \\[5mm]&=\color{#66f}{\large\ln\pars{k + 1}} \end{align}

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