Prove that $\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\cdots $

Question

Prove that

$$\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\frac{(m+1)^3 \ln^3 x}{1 \times 2 \times 3^2}+\cdots \infty $$

My Try:

I started with $$I=\int \frac{x^m}{\ln x}=\int \frac{x^{m+1} dx}{x \ln x}$$

Now by using parts taking $u=x^{m+1}$ and $v=\frac{1}{x \ln x}$ we get

$$I=x^{m+1} \times \ln(\ln x)-(m+1)\int x^m \ln (\ln x)dx$$

can i have any clue to proceed

Answer 1

With substitution $x=e^u$: \begin{align} \int \frac{x^m}{\ln x}dx &=\int\dfrac{1}{u}e^{u(m+1)}du \\ &=\int\dfrac{1}{u}\sum_{n=0}^\infty \dfrac{u^n(m+1)^n}{n!}du \\ &=\int\left(\dfrac{1}{u}+\sum_{n=1}^\infty \dfrac{u^{n-1}(m+1)^n}{n!}du\right) \\ &=\ln u+\sum_{n=1}^\infty \dfrac{u^n(m+1)^n}{n.n!} \\ &=\ln\ln x+\sum_{n=1}^\infty \dfrac{\ln^nx(m+1)^n}{n.n!} \end{align}

Answer 2

Since differentiating is usually easier than finding primitives, and you are given a candidate, the simplest way is to differentiate the right hand side. Write that as

$$f(x) = \ln (\ln x) + \sum_{n = 1}^{\infty} \frac{(m+1)^n(\ln x)^n}{n\cdot n!}.\tag{1}$$

The series converges locally uniformly on $(0,+\infty)$, so it can be differentiated term-by-term. This yields

\begin{align} f'(x) &= \frac{1}{x\ln x} + \sum_{n = 1}^{\infty} \frac{(m+1)^n(\ln x)^{n-1}}{n!\cdot x} \\ &= \frac{1}{x\ln x} + \frac{1}{x\ln x}\sum_{n = 1}^{\infty} \frac{(m+1)^n(\ln x)^n}{n!} \\ &= \frac{1}{x\ln x} \sum_{n = 0}^{\infty} \frac{(m+1)^n(\ln x)^n}{n!} \\ &= \frac{1}{x\ln x} e^{(m+1)\ln x} \\ &= \frac{x^{m+1}}{x\ln x} \\ &= \frac{x^m}{\ln x} \end{align}

for $x > 1$, so the right hand side is indeed a primitive of the integrand.