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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?

Sandra West
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1 Answers1

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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.

lord.didger
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