Need help with $\int \limits_{0}^{1} \frac{x^3-x^2}{\ln x} dx$

Question

Calculate the integral $$ \int \limits_{0}^{1} \frac{x^3-x^2}{\ln x} dx $$

Answer 1

Let $x:=\exp(-t)$. We have $$I:=\int_0^1\,\frac{x^3-x^2}{\ln(x)}\,\text{d}x=\int^{\infty}_{0}\,\frac{\exp(-3t)-\exp(-4t)}{t}\,\text{d}t=\int_0^\infty\,\exp\left(-\frac72t\right)\,\text{sinhc}\left(\frac{t}{2}\right)\,\text{d}t\,,$$ where $$\text{sinhc}(u):=\left\{ \begin{array}{cc} 1&\text{if }u=0\,, \\ \frac{\sinh(u)}{u}&\text{if }u\neq 0\,. \end{array} \right.$$ Observe that $$\text{sinhc}(u)=\frac12\,\int_{-1}^{+1}\,\exp(us)\,\text{d}s\,.$$ Consequently, $$I=\frac{1}{2}\,\int_0^\infty\,\exp\left(-\frac{7}{2}t\right)\,\int_{-1}^{+1}\,\exp\left(\frac{st}{2}\right)\,\text{d}s\,\text{d}t=\frac12\,\int_{-1}^{+1}\,\int_{0}^\infty\,\exp\Biggl(-\left(\frac{7}{2}-\frac{s}{2}\right)\,t\Biggr)\,\text{d}t\,\text{d}s\,,$$ where Fubini's Theorem is applied. Therefore, $$I=\frac{1}{2}\,\int_{-1}^{+1}\,\frac{1}{\left(\frac{7}{2}-\frac{s}{2}\right)}\,\text{d}s=\int_{-1}^{+1}\,\frac{1}{7-s}\,\text{d}s=\ln\left(\frac{7+1}{7-1}\right)=\ln\left(\frac{4}{3}\right)\,.$$

In general, if $a$ and $b$ are real numbers such that $\min\{a,b\}\geq -1$, then $$\int_0^1\,\frac{x^a-x^b}{\ln(x)}\,\text{d}x=\frac{\ln(1+a)-\ln(1+b)}{a-b}\,.$$ The proof is essentially the same as above.

Answer 2

Use substitution $x=e^{-u}$, then \begin{align} I &= \int_0^1\frac{x^3-x^2}{\ln x}dx \\ &= \int_\infty^0\frac{e^{-3u}-e^{-2u}}{-u}(-e^{-u})du \\ &= -\int_0^\infty\frac{e^{-4u}-e^{-3u}}{u}du \\ &= -\int_0^\infty\frac{1}{s+4}-\frac{1}{s+3}ds \\ &= \ln\dfrac{s+3}{s+4}\Big|_0^\infty \\ &= \color{blue}{\ln\dfrac{4}{3}} \end{align}