# Improper Integral $\int_{1/e}^1 \frac{dx}{x\sqrt{\ln{(x)}}}$

I need some advice on how to evaluate it. $$\int\limits_\frac{1}{e}^1 \frac{dx}{x\sqrt{\ln{(x)}}}$$ Thanks!

-
The integral is not well-defined. $\ln(1/e)=-1$ ... – David Mitra May 22 '13 at 17:02
Try a u-substitution with $u=\ln(x)$. – Andy Bromberg May 22 '13 at 17:03
so I need to change $\int\limits_\frac{1}{e}^1$ to? – Ofir Attia May 22 '13 at 17:04
@AndyBromberg if I`m doing it so I get $$\int \frac{dt}{\sqrt{t}}$$ but still I need to change the values of the integral right? – Ofir Attia May 22 '13 at 17:06
@DonAntonio: perhaps you mean $$\int_{1/e}^1\frac1{x\sqrt{|\log(x)|}}\mathrm{d}x$$ – robjohn May 22 '13 at 17:17

Here's a hint: $$\int_{1/e}^1 \frac{1}{\sqrt{\ln x}} {\huge(}\frac{dx}{x}{\huge)}.$$ If you don't know what that's hinting at, then you don't understand substitutions. It's all about the chain rule. The part in the gigantic parentheses becomes $du$.

-
I understand, but still my problem is the know when to change the limits, I get : $$\int^1_\frac{1}{e} \frac{dt}{\sqrt{t}}$$ – Ofir Attia May 22 '13 at 17:23
You have $t = \ln x$. When $x = 1/e$, then $t=\ln(1/e)=-1$. When $x=1$, then $t=\ln 1 = 0$. So you have $\displaystyle\int_{-1}^0 \frac{dt}{\sqrt{t}}$. Since you're talking about square roots of negative numbers, you have a question of how to make sense of those. One branch, maybe. – Michael Hardy May 22 '13 at 20:54

To recap all that happened in the comments section:

Based on the initial problem of$$\int\limits_\frac{1}{e}^1 \frac{dx}{x\sqrt{\ln{(x)}}}$$

We perform a u-substitution with $u=\ln{x}$ and $du=\frac{dx}{x}$. Also, the bounds are converted to $\ln\frac{1}{e}=-1$ and $\ln{1}=0$. So we have:

$$\int\limits_{-1}^{0} \frac{du}{\sqrt{u}}=2\sqrt{u}\big|_{-1}^{0}=0-2i=-2i$$

And that's the solution!

-