Recursion Formulas of $\int x^{\alpha}\ln x \ \text{dx}$ and $\int\frac{\ln^{\beta}x}{x} \ \text{dx}$ Suppose that I have recursion formulas of $\int x^{\alpha}\ln x \ \text{dx}$ and $\int\frac{\ln^{\beta}x}{x} \ \text{dx}$, and suppose that i found them(integration by parts),
$$\int x^{\alpha}\ln x \ \text{dx}=\frac{x^{{\alpha} + 1}}{{\alpha} + 1}  \big[\ln x - \frac{1}{{\alpha} + 1}\big] + C_1$$
and,
$$\int\frac{\ln^{\beta}x}{x} \ \text{dx}=\frac{\ln^{\beta + 1}x }{\beta + 1}+C_2$$
I have trouble showing that for $\alpha=-1$ and $\beta=1$ they have the same value(I cannot put $\alpha=-1$)
What to do?  

For $\alpha=-1$, we have
\begin{align}
\int x^{-1} \ln(x) dx & = \overbrace{\int \dfrac{\ln(x)}x dx = \int t dt}^{t = \ln(x)} = \dfrac{t^2}2 + \text{constant}\\
& = \dfrac{\ln^2(x)}2 + \text{constant} = \dfrac{\ln^{1+1}(x)}{1+1} + \text{constant}
\end{align}
The expression you have is
\begin{align}
\int x^{\alpha} \ln(x) dx & = \dfrac{x^{\alpha+1} \ln(x)}{1+\alpha} - \dfrac{x^{\alpha+1}}{(1+\alpha)^2} + \text{constant}
\end{align}
You cannot directly take the limit as $\alpha \to -1$. But note that you can draw some constants out from the constant term to help you.
\begin{align}
\int x^{\alpha} \ln(x) dx & = \dfrac{(x^{\alpha+1} - 1) \ln(x)}{1+\alpha} + \dfrac{\ln(x)}{1+\alpha}+ \dfrac{1-x^{\alpha+1}}{(1+\alpha)^2} \underbrace{- \dfrac1{(1+\alpha)^2}+ \text{constant}}_{\text{new constant}}\\
& = \dfrac{(x^{\alpha+1} - 1) \ln(x)}{1+\alpha} + \dfrac{\ln(x)}{1+\alpha}+ \dfrac{1-x^{\alpha+1}}{(1+\alpha)^2} + \text{constant} \,\,\,\, (\spadesuit)
\end{align}
Now note that
$$x^{1+\alpha} = \exp((1+\alpha) \ln(x)) = 1 + (1+\alpha) \ln(x) + \dfrac{(1+\alpha)^2}2 \ln^2(x) + \mathcal{O}((1+\alpha)^3)$$
Hence,
\begin{align}
\dfrac{1- x^{1+\alpha}}{1+\alpha} & = - \ln(x) - \dfrac{1+\alpha}2 \ln^2(x) + \mathcal{O}((1+\alpha)^2)\\
\ln(x) + \dfrac{1- x^{1+\alpha}}{1+\alpha} & = - \dfrac{1+\alpha}2 \ln^2(x) + \mathcal{O}((1+\alpha)^2)\\
\dfrac{\ln(x)}{1+\alpha} + \dfrac{1- x^{1+\alpha}}{(1+\alpha)^2} & = - \dfrac{\ln^2(x)}2 + \mathcal{O}((1+\alpha))
\end{align}
Plug this in $(\spadesuit)$ to get
$$\int x^{\alpha} \ln(x) = \ln^2(x) + \dfrac{1+\alpha}2 ln^3x + ln x\,\mathcal{O}((1+\alpha)) - \dfrac{\ln^2(x)}2 + \mathcal{O}((1+\alpha)) = \dfrac{\ln^2(x)}2 + \dfrac{1+\alpha}2 ln^3x + (ln x + 1)\,\mathcal{O}((1+\alpha))$$
Now letting $\alpha \to -1$, we get that
$$\lim_{\alpha \to -1} \int x^{\alpha} \ln(x) = \lim_{\alpha \to -1} \dfrac{\ln^2(x)}2 + \mathcal{O}((1+\alpha)) = \dfrac{\ln^2(x)}2$$
which matches with our original integral.
