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so what this means is if $f(x)$ is the indefinite integral of $x^x dx$ then what would the indefinite integral of $x^{1/x}$ be in terms of $x$ and $f(x)$

Ekadh Singh - Reinstate Monica
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  • Do you mean $f$ is an antiderivative of $x \mapsto x^x$ ? – Raito Jun 29 '19 at 22:03
  • Yay, you can't write "$x^x=f(x)$" here. You have to write "$x^x$ is $f(x)$". – TonyK Jun 30 '19 at 01:11
  • IN the bounty description I messed up when I wrote ox I meant x – Ekadh Singh - Reinstate Monica Jul 02 '19 at 23:23
  • $F(x) = \int_0^x t^t \mathrm{d}t$ is an antiderivative of $x^x$, meaning that $F(x) + C$ is the indefinite integral of $x^x$. Note that it isn't so much a function as it is an entire family of functions. It sounds extra impossible to express the family of functions $\int x^{1/x} , \mathrm{d}x$ in terms of the family of functions $\int x^x , \mathrm{d}x$ in a way that handles the constant of integration, and doesn't simply involve differentiating back to $x^x$. – Theo Bendit Jul 03 '19 at 00:08
  • So are you saying that the indefinite integral of x^x Dax is the integral between 0 and x of t^t dt?? – Ekadh Singh - Reinstate Monica Jul 03 '19 at 12:52
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    @Yay It's worse! It's $\int_0^x t^t , \mathrm{d}t$ + C. The indefinite integral has a constant of integration. There is no one function that is the indefinite integral of $x^x$ (or any function). Instead, you get a whole family of functions, differing by a constant. Despite appearances, there is no one antiderivative (corresponding to a single value of $C$) that is more "natural" than another! Writing $\int x^{1/x} , \mathrm{d}t$ in terms of one antiderivative seems hard enough, but writing it in terms of a whole family of functions seems unreasonable. – Theo Bendit Jul 03 '19 at 15:32

3 Answers3

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Partial answer

Let be $g : x \mapsto \int_1^x t^{1/t} \textrm{d}t$ and $h : x \mapsto \int_0^x t^{-t} \textrm{d}t$.

Thus, if we do the change of variable $u = 1/t$ in the first integral, we have, for all $x \geq 1$:

\begin{align*} \int_1^x t^{1/t} \textrm{d}t & = -\int_{1/x}^1 u^{-(u + 2)} \textrm{d}u \\ & = -\int_{1/x}^1 1/u^2 u^{-u} \textrm{d}u \\ & = -\left(\left[h(u)/u^2\right]_{1/x}^1 + 2\int_{1/x}^1 \dfrac{h(u)}{u^3} \textrm{d}u\right) \\ & = \left(x^2 h(1/x) - h(1)\right) - 2\int_{1/x}^1 \dfrac{h(u)}{u^3} \textrm{d}u \end{align*}

We then have:

\begin{equation*} g(x) = (x^2h(1/x) - h(1)) - 2\int_{1/x}^1 h(u)/u^3 \textrm{d}u \end{equation*}

We can continue this by 2nd MVT theorem, as $h$ is continuous (even differentiable!) and $u \mapsto 1/u^3$ non negative and continuous, there is some $c_x \in [1/x, 1]$ such that:

\begin{equation*} \int_{1/x}^1 \dfrac{h(u)}{u^3} \textrm{d}u = -\dfrac{h(c_x)}{2}\left(1 - x^2\right) \end{equation*}

Finally: \begin{equation*} g(x) = x^2 (h(1/x) - h(c_x)) - h(1) + h(c_x) \end{equation*}

Now, I'm unsure if there is an easy relation to $f : x \mapsto \int_{0}^x t^t \textrm{d}t$, if you try to apply some techniques, you will see that it's difficult to get $-t$.

Raito
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I think you mean if: $$f(x)=\int_0^xt^tdt$$ then what is: $$\int t^{1/t}dt$$ Which I do not think is possible to define just using $x$ and $f(x)$

Or if you mean: $$f(x)=x^x$$ then what is: $$x^{1/x}$$ in which case see that: $$f(1/x)=\frac{1}{x^{1/x}}$$ $$\frac{1}{f(1/x)}=x^{1/x}$$ Hence: $$x^x=f(x),x^{1/x}=\frac{1}{f(1/x)}$$

Henry Lee
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We have $\displaystyle f(x)=\int x^x\,dx.$ If we have good domain issues, we can compute the following: \begin{align*} f(x)&=\int x^x\,dx \\ f'(x)&=x^x \\ \left(f'(x)\right)^{1/x}&=x \\ \left(f'(x)\right)^{1/x^2}&=x^{1/x} \\ \int\left(f'(x)\right)^{1/x^2}\,dx&=\int x^{1/x}\,dx. \end{align*} You'd need to have $f'(x)$ behave well, as the exponent is fractional, but this is a formal manipulation that could give you some notion of the answer.

Also note that this technically doesn't answer the question, since I have $f'(x)$ instead of $f(x);$ however, this could improve your intuition.

Adrian Keister
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