Evaluate $\int^{1/2}_0\frac{1}{x(\ln(1/x))^2}\text{d}x$

Question

I need to evaluate $\displaystyle\int^{1/2}_0\frac{1}{x(\ln(1/x))^2}\text{d}x$

So the integral is $\displaystyle\int\frac{1}{-x\ln(x)^2}\text{d}x$

I let $\displaystyle u=\ln(x)$ and then $\displaystyle\text{d}u=\frac{1}{x}\text{d}x$ so $\displaystyle -\int\frac{x}{xu^2}\text{d}u=-(-\frac{1}{u})$

So $\displaystyle\int^{1/2}_0\frac{1}{x(\ln(1/x))^2}\text{d}x=\frac{1}{\ln(1/2)}?$

Answer 1

You missed a minus sign. Note that

$$\begin{align}\left(\log(1/x)\right)^2&=\left(-\log(x)\right)^2\\\\ &=(-1)^2(\log(x))^2\\\\ &=(\log(x))^2\\\\ &\ne -(\log(x))^2 \end{align}$$

The rest of your work was fine and the result is $1/\log(2)$.

Answer 2

It is almost correct, you just missed a sign. It should be $$\int^{1/2}_0\frac{1}{x(\ln(1/x))^2}dx=\int^{1/2}_0\frac{1}{x(\ln(x))^2}dx=\left[-\frac{1}{\ln(x)}\right]_0^{1/2}=\frac{1}{\ln(2)}.$$

Answer 3

Since squaring is even, it's $\int_0^{1/2}\frac{dx}{x\ln^2x}$. With $u=-\ln x$, this is $\int_{\ln 2}^\infty\frac{du}{u^2}=\frac{1}{\ln2}$.