Definite integration integration by parts

Question

Evaluate the integral $$\int_0^1\frac{x^7-1}{\log (x)}\,dx $$

[1]: https://i.stack.imgur.com/lcK2p.jpgplz I'm trying to solve this definite integral since 2 hours. Please, I need help on this.

Answer 1

Note that using $\int_0^1 x^t \,dt=\frac{x-1}{\log(x)}$ we can write

$$\begin{align} \int_0^1 \frac{x^7-1}{\log(x)}\,dx&=\int_0^1 (x^6+x^5+x^4+x^3+x^2+x+1)\left(\int_0^1 x^t\,dt\right)\,dx\tag 1\\\\ &=\int_0^1\int_0^1 (x^{t+6}+x^{t+5}+x^{t+4}+x^{t+3}+x^{t+2}+x^{t+1}+x^t)\,dx\,dt\tag2 \\\\ &=\int_0^1 \left(\frac{1}{t+7}+\frac{1}{t+6}+\frac{1}{t+5}+\frac{1}{t+4}+\frac{1}{t+3}+\frac{1}{t+2}+\frac{1}{t+1}\right)\,dt\\\\ &=\log(8) \end{align}$$

where the Fubini-Tonelli Theorem guarantees the legitimacy of interchanging the order of integration in going from $(1)$ to $(2)$.

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Note that if we first enforce the substitution $x^7\to x$, we obtain

$$\begin{align} \int_0^1 \frac{x^7-1}{\log(x)}\,dx&=\int_0^1 x^{1/7-1}\left(\frac{(x-1)}{\log(x)}\right)\\\\ &=\int_0^1 x^{1/7-1}\left(\int_0^1 x^t\,dt\right)\,dx\\\\ &=\int_0^1\int_0^1 x^{t+1/7-1} \,dx\,dt\\\\ &=\int_0^1 \frac{1}{t+1/7}\,dt\\\\ &=\log(8) \end{align}$$

as expected!

Answer 2

You can use feynman approach Consider \begin{align} f(n)=\int_0^1 \frac{x^n-1}{\log(x)}\,dx& \end{align} Then \begin{align} f'(n)=\int_0^1 \frac{x^n\log(x)}{\log(x)}\,dx& \end{align} So \begin{align} f'(n)= \frac{1}{n+1} \end{align} \begin{align} f(n)=\log(n+1)+c \end{align} For c \begin{align} f(0)=0 , c=0 \end{align} \begin{align} f(7)=\int_0^1 \frac{x^7-1}{\log(x)}\,dx&=\log(8) \end{align}

Answer 3

This can be written as a Frullani Integral: $$ \begin{align} \int_0^1\frac{x^7-1}{\log(x)}\,\mathrm{d}x &=\int_0^\infty\frac{e^{-u}-e^{-8u}}{u}\,\mathrm{d}x\tag{1}\\ &=\lim_{\epsilon\to0}\int_\epsilon^\infty\frac{e^{-u}-e^{-8u}}{u}\,\mathrm{d}x\tag{2}\\ &=\lim_{\epsilon\to0}\left[\int_\epsilon^\infty\frac{e^{-u}}{u}\,\mathrm{d}x-\int_{8\epsilon}^\infty\frac{e^{-u}}{u}\,\mathrm{d}x\right]\tag{3}\\ &=\lim_{\epsilon\to0}\int_\epsilon^{8\epsilon}\frac{e^{-u}}{u}\,\mathrm{d}x\tag{4}\\ &=\lim_{\epsilon\to0}\int_\epsilon^{8\epsilon}\frac{1+O(u)}{u}\,\mathrm{d}x\tag{5}\\[3pt] &=\lim_{\epsilon\to0}\left(\log(8)+O(\epsilon)\right)\tag{6}\\[6pt] &=\log(8)\tag{7} \end{align} $$ Explanation:
$(1)$: $x=e^{-u}$
$(2)$: write as a limit near $0$
$(3)$: substitute $u\mapsto\frac{u}8$
$(4)$: combine integrals
$(5)$: $e^u=1+O(u)$ for $u$ near $0$
$(6)$: integrate
$(7)$: take the limit

Answer 4

hint for the convergence

$$f : x\mapsto \frac {x^7-1}{\log (x)} $$ is continuous at $(0,1) $ , and is locally integrable.

Near $0$, $f $ is bounded since $\lim_{0^+}f (x)=0$ thus $$\int_0^\frac 12 f (x)dx $$ is convergent.

Near $1$, $$\log (x)\sim x-1$$ and $$\lim_{1^-}f (x)=6$$ thus $\int_\frac 12^1f (x)dx $ converges.

finally $\int_0^1f (x)dx $ is convergent.