$f(n) = \int_{1}^{n} n^{x^{-1}}dx$. $\frac{df}{dn}$?

Question

Lets start with the domain $1 \leq n \in \mathbb{R}$:

$f(n) = \int_{1}^{n} n^{x^{-1}}dx$

$\frac{df}{dn}$ should be sort of logarithmic and probably not interesting.?

What is it, though?

What have I tried? To see how far it is from something interesting, and maybe more..

Answer 1

METHODOLOGY $1$: EVALUATION OF $\displaystyle f(n)$ AND DIRECT DIFFERENTIAION

EVALUTING $\displaystyle f(n)$:

We can write the integral of interest in terms of the Exponential Integral $\text{Ei}(x)$.

Note that we have

$$\begin{align} f(n)&=\int_1^n n^{1/x}\,dx\tag1\\\\ &=\int_{1/n}^1 \frac{n^x}{x^2}\,dx\\\\ &=\int_{1/n}^1 \frac{e^{\log(n)x}}{x^2}\,dx\\\\ &=\log(n)\int_{\log(n)/n}^{\log(n)} \frac{e^x}{x^2}\,dx\\\\ &=\log(n)\left(\text{Ei}(\log(n))-\text{Ei}(\log(n^{1/n}))\right)+n(n^{1/n}-1)\tag 2 \end{align}$$

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DIFFERENTIATING $\displaystyle f(n)$:

To determine $f'(n)$ we begin with $(2)$ and write

$$\begin{align} f'(n)&=\frac1n\left(\text{Ei}(\log(n))-\text{Ei}(\log(n^{1/n}))\right)\\\\ &+\log(n)\left(\frac{1}{\log(n)}-\frac{n}{\log(n)}\frac{dn^{1/n}}{dn}\right)\\\\ &+(n^{1/n}-1)+n\frac{dn^{1/n}}{dx}\\\\ &=n^{1/n}+\frac1n\left(\text{Ei}(\log(n))-\text{Ei}(\log(n^{1/n}))\right)\tag 3 \end{align}$$

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METHODOLOGY $2$: APPLYING LEIBNIZ'S RULE

Using Leibniz's Rule, we differentiate $(1)$ to obtain

$$\begin{align} f'(n)&=n^{1/n}+\int_1^n \frac{n^{1/x-1}}{x}\,dx\\\\ &=n^{1/n}+\frac1n \int_{1/n}^1 \frac{n^x}{x}\,dx\\\\ &=n^{1/n}+\frac1n \int_{\log(n)/n}^{\log(n)}\frac{e^x}{x}\,dx\\\\ &=n^{1/n}+\frac1n\left(\text{Ei}(\log(n))-\text{Ei}(\log(n^{1/n}))\right) \end{align}$$

which agrees with the result reported in $(3)$!