# For which parameters is the following function defined on the reals, and how to integrate it?

The value $a=1$ seems to be the only one for which

$$f_a(x):=\frac{1}{\mathrm e^x-x^a}$$

is defined for all $\mathbb R$. Can someone shine a light on why this value is special? What fails in the other cases?

To my surprise, that function $$f_1(x)=\frac{1}{\mathrm e^x-x}=\frac{1}{1+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots},$$ which is $\approx-\frac{1}{x}$ on the left and $\approx \mathrm e^{-x}$ on the right hand side of the $y$-axis, seems to be incredibly hard to integrate. I think $$\int \frac{1}{\mathrm e^x-x}\mathrm dx$$ might be non-elementary. The series expansion should have a somewhat compact form, though I don't know what it is. Can we integrate $f_1(x)$, or is there at least a faster method to compute it than integrate the terms of it's tailor expansion?

(These are technically two questions, but the first is motivated by me solving the second one.)

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The value$a=\frac13$ allows $f_a$ to be defined on $\Bbb R$, under the reasonable interpretation of $x^{1/3}$ as $-\sqrt[3]|x|$ for $x<0$. – John Bentin Oct 14 '13 at 18:40

Your function is only defined on all of ${\mathbb R}$ if $a = p/q$ for relatively prime $p/q$ where $q$ is odd. (Otherwise you have to take square-roots of negative numbers and raise them to odd powers, which isn't possible over the reals). If $a < 0$ it is easy to see the function $e^x = x^a$ has a solution, because $e^x - x^a$ is negative as $x \to 0^+$, but the function is positive for $x > 1$. If $a > 0$ where $x^a$ is even then you get $e^x - x^a$ is negative for large negative $x$, while positive for large positive $x$, so $e^x = x^a$ again has a solution. On the other hand, if you have $1 > a > 0$ where $x^a$ is odd, then $e^x - x^a$ will always be positive for negative $x$. For non-negative $x$, you can note that $e^x - x^a$ will still always be positive because the minimum occurs when the derivative is zero, i.e. $e^x = ax^{a-1}$, and in this case at the minimum the function value is $ax^{a-1} - x^a = (a-x)(x^{a-1})$ , and we must have $a > x$ because if $x \leq a$ then $ax^{a-1} \leq a^a < 1 \leq e^x$.
So in summary, if $0 < a < 1$ such that $x^a$ is odd, then your function is well-defined. In particular $a= 1/3$ works as noted in a comment.