# Evaluate log improper integral $\int_{a}^{\infty} -\frac{1}{c (1-x) x}$.

I'm trying to evaluate

$$\int_{a}^{\infty} -\frac{1}{c (1-x) x}$$ where $c, a$ are constants.

Mathematica gives:

$$-\dfrac{\log \left(\dfrac{a-1}{a}\right)}{c}$$

where I assume that $a > 1$. How does one get this result? Since, the indefinite integral evaluates to $(\log(1-x) - \log(x)) / c$, and then taking the limit as $x \to \infty$ gives $i \pi$.

Thanks.

HINT: you will get $$\frac{\log(1-x)-\log(x)}{c}$$ and it must be $$0<x<1$$

• I think you actually get (Log[1-x] - Log[x]) / c, and not divided by $x$. – Thomas Moore Sep 4 '17 at 17:16
• sorry for my typo,just corrected – Dr. Sonnhard Graubner Sep 4 '17 at 17:17
• No problem. But, where do you go from here. I got up to this step. – Thomas Moore Sep 4 '17 at 17:20
• i think we must write $$\frac{\log|1-x|-\log|x|}{c}$$ since $$\int\frac{1}{x}dx=\ln|x|+C$$ – Dr. Sonnhard Graubner Sep 4 '17 at 17:24

$\int_a^{+\infty}-\frac{1}{c(1-x)x}dx=\lim_{t \to +\infty}\int_a^t-\frac{1}{c(1-x)x}dx$

Now $$-\frac{1}{(1-x)x}=\frac{1}{(x-1)x}$$

Also $$\frac{1}{(x-1)x}=\frac{1}{x-1}-\frac{1}{x}$$

Thus solving the integral, you have to compute :$$\lim _{t \to +\infty} \frac{1}{c}(-\log{\frac{a-1}{a}}+\log{\frac{t-1}{t}})$$

But $$\lim_{t \to +\infty} \log{\frac{t-1}{t}}=0$$

• Hi. Can you plz explain how you got the second line? Because I thought integral of 1/x-1 = log(x-1). – Thomas Moore Sep 4 '17 at 17:19
• oh. of course. Thanks! – Thomas Moore Sep 4 '17 at 17:21
• @ThomasMoore i edited and i also use the property of the difference of logarithms ..I hope it helps now. – Marios Gretsas Sep 4 '17 at 17:24

Apart from constant $-\dfrac1c$, with substitution $x=\dfrac1u$ we have $$I=\int_{a}^{\infty} \frac{1}{ (1-x) x}dx=\int_0^{\frac1a} \frac{1}{u-1}du$$ $0\leqslant u\leqslant \dfrac1a$ so with assumption $a>1$, $-1\leqslant u-1 \leqslant \dfrac1a-1<0$, then $$I=-\int_0^{\frac1a} \frac{1}{1-u}du=\ln\left(1-\dfrac1a\right)$$