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How can we calculate this integral:

$$I=\int\limits_0^1\frac{x^n-1}{\ln(x)}\,\mathrm dx$$

I believe that integral is equal to $\ln(n+1)$, but I don't lnow how to prove it.

Olivier Oloa
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mathlover
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5 Answers5

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One may observe that $$\frac{x^n-1}{\ln x}=n\int_{0}^{1}x^{nu} du$$ plugging it in your initial integral, applying Fubini's theorem, gives $$ \int_0^1\frac{x^n-1}{\ln x}dx= n\int_0^1\!\!\frac{1}{nu+1}du=\int_0^1\frac{(nu+1)'}{nu+1}du=\ln(n+1). $$

Olivier Oloa
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  • @OliviwrOloa I took the liberty of fixing a typo - a missing $n$ in the exponent of the first integrand. I hope that you don't mind my doing so. – Mark Viola May 02 '15 at 20:58
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If we replace $x$ with $e^{-t}$ we are left with: $$ I = \int_{0}^{+\infty}\frac{e^{-t}-e^{-(n+1)t}}{t}\,dt = \log(n+1) $$ by Frullani's theorem.

Mark Viola
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Jack D'Aurizio
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  • You would need to prove that the integral converges though. – Adhvaitha May 02 '15 at 20:19
  • @user17762: that is quite trivial since the integrand function is positive, bounded on $[0,1]$ and less than $e^{-t}$ on $[1,+\infty)$. – Jack D'Aurizio May 02 '15 at 20:27
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    @JackD'Aurizio I answered a problem or two on Frullani ... development and application. And here is yet another! It's a really neat equality. – Mark Viola May 02 '15 at 21:02
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    I corrected the dummy differential in the integral from $x$ to $t$ in order to match the dummy variable of the integrand. – Mark Viola May 02 '15 at 21:45
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Let: $$I(n)=\int_0^1\dfrac{x^n-1}{\ln x}dx$$ Then: \begin{align}I'(n)&=\dfrac{d}{dn}\int_0^1\dfrac{x^n-1}{\ln x}dx=\int_0^1\dfrac{\partial}{\partial n}\left[\dfrac{x^n-1}{\ln x}\right]dx\\ &=\int_0^1 x^n dx=\left.\dfrac{x^n+1}{n+1}\right\vert_0^1\\ &=\dfrac{1}{n+1} \end{align} Therefore: $$I(n)=\int I'(n)=\ln(n+1)$$

Demosthene
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2

Hint: $(a^x)'=a^x\cdot log(a)$

Lucas
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We have $$x^a = e^{a\log x} = \sum_{k=0}^{\infty} \dfrac{(a \log x)^k}{k!}$$ Hence, $$\dfrac{x^a-1}{\log(x)} = \sum_{k=1}^{\infty} \dfrac{a^k \log^{k-1}(x)}{k!}$$ Make use of the fact that $$\int_0^1 \log^m(x) dx = (-1)^m m!$$ Hence, we obtain $$\dfrac{x^a-1}{\log(x)} = \sum_{k=1}^{\infty} \dfrac{a^k}{k!} (-1)^{k-1} (k-1)! = \sum_{k=1}^{\infty} (-1)^{k-1} \dfrac{a^k}k = \log(1+a)$$

Note that the above derivation is only valid when $a \in (-1,1)$. For $a\geq1$, it is easy to show by similar method that $$I(a) - I(a-1) = \log(1+1/a)$$ Since $I(a) = \log(1+a)$ on the interval $[0,1]$, we can now conclude that $$I(a) = \log(1+a)$$ for all $a \geq -1$.

Adhvaitha
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