Evaluating $\int_0^1 \frac{x^3-1}{\ln x}dx$

Question

The question is to evaluate $$\int_0^1 \frac{x^3-1}{\ln x}dx$$

I tried to put $t=\ln x$ so $dt=\frac1xdx$.Hence I could rewrite the integral as $$\int_{-\infty}^0 \frac{e^{3x}-1}{x}e^xdx$$I couldnot further simplify the integral.Any ideas?Thanks.

Answer 1

$$\displaystyle \int^{1}_{0}\frac{x^3-1}{\ln x}dx = \int^{1}_{0}\left(\int^{3}_{0}x^{ t}dt\right)dx = \int^{3}_{0}\left(\int^{1}_{0}x^{t}dx\right)dt$$

So $$\displaystyle \int^{3}_{0}\left(\frac{x^{t+1}}{t+1}\right)\bigg|_{0}^{1}dt = \int^{3}_{0}\frac{1}{t}dt = \ln(t+1)\bigg|_{0}^{3} = \ln (4)$$

Answer 2

Note that substituting $x=e^{-t}$, your integral becomes $$\int_0^1 \frac{x^3-1}{\ln x}\mathrm{d}x=\int_{0}^{\infty} \frac{e^{-t}-e^{-4t}}{t} \mathrm{d} t$$

As $\ln x =-t$. Now use Frullani's Integral to get that $$\int_{0}^{\infty} \frac{e^{-t}-e^{-4t}}{t} \mathrm{d} t=\ln \frac{-4}{-1}=\ln 4$$

As corroborated here. We are done!

Answer 3

As shown here, it's easy enough to take a more general problem:

$$f(t)=\int_0^1\frac{x^t-1}{\ln(x)}\ dx$$

Differentiate with respect to $t$ and you'll get

$$f'(t)=\int_0^1x^t\ dx=\frac1{t+1}$$

Integrate both sides from zero to one to reveal

$$f(t)=\ln(t+1)$$

$$\int_0^1\frac{x^t-1}{\ln(x)}\ dx=\ln(t+1)$$

With $t=3$ being your integral.