How would you do this integral?

Question

I'm currently preparing a little conference on the Feynman integration method for my high school, and I chose a particular example to apply this method: $$\int_{0}^{1}\frac{x^2-1}{\log(x)} \,\mathrm{d}x$$ This integral is really easy to compute with the Feynman method, that's why I chose it. But the thing is, I would like to show my students how could we do it using another method so if you are motivated to try this integral using sub, IBP or anything else I would be curious to see.

Answer 1

$$I=\int_{0}^{1}\frac{x^2-1}{log(x)}dx$$

Put in $$t=-log(x)$$ $$-e^{-t}dt=dx$$ We get,

$$I=\int_{\infty}^{0}\frac{e^{-3t}-e^{-t}}{t}\,dt$$

$$I=\int_{0}^{\infty}\frac{e^{-t}-e^{-3t}}{t}\,dt$$

The above integral is a Frullani Integral,

$$I=ln(3)$$

Frullani Integral;

$$I=\int_{0}^{\infty}\frac{f(ax)-f(bx)}{x}\,dx=(f(\infty)-f(0))ln\left(\frac{a}{b}\right)$$

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Using the Feynman Technique;

$$I(\alpha)=\int_0^1\frac{x^\alpha-1}{\log(x)}\;dx$$$$\frac{\partial}{\partial\alpha}I(\alpha)=\int_0^1\frac{\partial}{\partial\alpha}\frac{x^\alpha-1}{\log(x)}\;dx=\int_0^1x^\alpha\;dx=\frac{1}{\alpha+1}$$ $$\displaystyle I(\alpha)=\log(\alpha+1)+C$$ Using $I(0)=0$ $$I(\alpha)=\log(\alpha+1)$$ $$I(2)=\log(3)$$

Answer 2

$$ I = \int_{0}^{1} \frac{x^{n}-1}{\ln(x)}\, \mathrm{d}x \overset{\color{blue}{t = -\ln(x)}}{=} \int_{0}^{\infty} \frac{e^{-t}}{t}\left(1- e^{-nt}\right)\, \mathrm{d}t $$ Since $e^{x} = \sum_{k \ge 0}\frac{x^k}{k!} = 1+\sum_{k \ge \color{purple}{1}}\frac{x^k}{k!} $ then \begin{align*} I & = -\int_0^{\infty} \frac{e^{-t}}{t}\sum_{k \ge 1}\frac{(-nt)^k}{k!}\, \mathrm{d}t\\ & = -\sum_{k \ge 1}\frac{(-n)^k}{k!} \int_0^{\infty} t^{k-1}e^{-t}\, \mathrm{d}t\\ \end{align*} For positive integer $m$, successive aplications of integration by parts leads to $\int_{0}^{\infty} t^m e^{-t}\, \mathrm{d}t = m!$. This argument can be formalized using induction. Thus $$ I = - \sum_{k \ge 1} \frac{(-n)^k}{k!}(k-1)! = - \sum_{k \ge 1} \frac{(-n)^k}{k} = \ln(n+1) $$ where on the last step you recognize the Taylor series for $\ln(1-x) = -\sum_{k \ge 1} \frac{x^k}{k}$.

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It is worth noting that the key evaluation $\int_0^{\infty} t^{m}e^{-t}\, \mathrm{d}t = m!$ is in reality just a special case of the Gamma function for integer inputs. Proofs concerning the Gamma function are usually done using double integrals/Feynman's trick, and this would arguably make the evaluation simpler in this case as well, but since OP mentions that they already know evaluations using these methods I attempted to keep it a bit more grounded in high school knowledge.