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Set an $\epsilon\in(0,1)$. How can we solve the integral $\int_\epsilon^1 e^{x^n\log x }{\rm d }x$ for all $n\in\mathbb{N}$?

My attempt. Set $I_n=\int_\epsilon^1 e^{x^n\log x }{\rm d }x$. I tried a recursive procedure to express $I_{n}$ in terms of $I_{n-1}$: \begin{align} I_n=& \int_\epsilon^1 e^{x^n\log x }{\rm d }x\\ =& \int_\epsilon^1 D_x(x)e^{x^n\log x }{\rm d }x\\ =& xe^{x^n\log x }\Big|_\epsilon^1-\int_\epsilon^1 xD_x(e^{x^n\log x }){\rm d }x\\ \end{align}

But it did not work. So I tried to use some feature of symbolic integration: (see this answer).

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Elias Costa
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  • I doubt there is a nice closed form expression for this integral. You might find a nice series solution though if that is of interest. – Winther Nov 02 '14 at 15:30
  • @Winther I really do not know. Maybe for an appropriate value of $\epsilon>0$ we could discover a serie that converges to the same value of the integral. – Elias Costa Nov 02 '14 at 15:34
  • If you take $\epsilon = 0$ then $\int_0^1 e^{x^n\log x}dx = \sum_{k=0}^\infty \frac{(-1)^k}{(1 + n k)^{k + 1}}$ – Winther Nov 02 '14 at 15:37
  • @Winther, thanks. But how you got this result? – Elias Costa Nov 02 '14 at 15:39
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    Expand $e^{x^n\log x}$ in a Taylor series. Then you get sums of integrals $\int x^m \log^k x dx$ which can be solved by making the substitution $t = -\log x$ and noticing that the integral you then get is nothing but the Gamma-function. – Winther Nov 02 '14 at 15:41
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    Since $a^{\log b}=b^{\log a}$, your expression seems to be an attempt at generalizing the Sophomore's dream. – Lucian Nov 02 '14 at 16:19
  • @Lucian +1 for your very interesting link. – Elias Costa Nov 02 '14 at 16:37
  • @Winther Around which point $x_o$ between $0$ and $1$ the Taylor expansion of $e^{x^n\log x}$ is centered? – Elias Costa Nov 02 '14 at 17:33
  • Around $x=0$: $x^{x^n} = e^{x^n\log x} = \sum_{k=0}^\infty \frac{1}{k!} x^{nk}(\log x)^k$. btw you might worry about the validity at $x=0$, but since $\lim_{x\to 0} x^{nk}(\log x)^k = \lim_{x\to 0} x^{(n-1)k}(x \log x)^k = 0$ everything is well defined for all $x\geq 0$. – Winther Nov 02 '14 at 17:38
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    The function is not defined as written, but the limit exist so if we define $f(x) = e^{x^n \log x}$ for $x>0$ and $f(x) = 1$ for $x=0$ then this is a continious function. The Taylor expansion for this function is identical as for the previous function for all $x>0$. – Winther Nov 02 '14 at 17:42
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    That was badly worded by me. You should start by defining $f(0)=1$ and then work with this modified function all the way. Then the Taylor series is well defined and gives the desired result. Note that for the purpose of integration it does not matter what you define $f(0)$ to be (as long as it is finite) since one point cannot alter the value of the integral (in technical terms one point has measure zero). – Winther Nov 02 '14 at 17:51
  • @Winther Thank you for detailing your observations so readily. I think I can work with this idea and maybe get some interesting partial results. – Elias Costa Nov 02 '14 at 17:53
  • @Winther Guiding me through the comments got a double set equal to the integral. – Elias Costa Nov 03 '14 at 12:49

1 Answers1

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This answer is based on the comments of Lucian and especially Winther. Expand $$ e^{(x^n \log x)} = \sum_{k=0}^\infty \frac{(x^{n})^k(\log x)^k}{k!} $$ Therefore we have $$ \int_\epsilon^1 e^{(x^n \log x)}\,\mathrm{d}x = \int_\epsilon^1 \sum_{n=0}^\infty \frac{(x^{n})^k(\log x)^k}{k!} \,\mathrm{d}x $$ By uniform convergence of the power series, we may interchange summation and integration $$ \int_\epsilon^1 e^{(x^n \log x)}\,\mathrm{d}x = \sum_{n=0}^\infty \int_\epsilon^1 \frac{x^{nk}(\log x)^k}{k!}\,\mathrm{d}x. $$ According to Wikpedia $\int x^m (\ln x)^k\,\mathrm{d}x = \frac{x^{m+1}}{m+1} \cdot \sum_{i=0}^k (-1)^i \frac{(k)_i}{(m+1)^i} (\ln x)^{k-i}$ for all $m\neq -1$ where $(k)_i$ denotes the falling factorial. Set $m=kn$. Then \begin{align} \int_\epsilon^1 \frac{x^{kn}(\ln x)^k}{k!}\,\mathrm{d}x =& \sum_{i=0}^k (-1)^i \frac{1}{k!}\cdot\frac{x^{kn+1}}{(km+1)} \cdot \frac{(k)_i}{(km+1)^i} (\ln x)^{k-i}\Bigg|_\epsilon^1 \\ =&\sum_{i=0}^k (-1)^{i+1} \frac{1}{k!}\cdot\frac{x^{kn+1}}{(km+1)} \cdot \frac{(k)_i}{(km+1)^i} (\ln \epsilon)^{k-i} \end{align} and finally $$ \int_\epsilon^1 e^{x^n\log(x)}\,\mathrm{d}x = \sum_{k=0}^\infty \sum_{i=0}^k (-1)^{i+1} \frac{1}{k!} \cdot \frac{x^{kn+1}}{(km+1)} \cdot \frac{(k)_i}{(km+1)^i} (\ln \epsilon)^{k-i} $$

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