Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the minimum value of this integral: For what value of $k > 1$ is \[ \int_k^{k^2} \frac 1x \log\frac{x-1}{32}\; dx \] minimal?

After applying Newton-Leibniz, I got $k = 3$ and then did 2nd derivative test, it gave me positive result, thus 3 is the answer but I want to know if there's a smarter/slicker way to do?

share|cite|improve this question
That green hurts my eyes. You can use $\TeX$ on this site, enclosed in single dollar signs for inline formulas and in double dollar signs for displayed equations. – joriki Jul 12 '12 at 13:34
Sorry sir. I'll take care of it from next time. – Hyperbola Jul 12 '12 at 13:35
Two answers, but I'm the only person who's up-voted the question so far. – Michael Hardy Jul 12 '12 at 18:02
up vote 6 down vote accepted

Let $I:(1, \infty) \to \mathbb{R}$ be defined by $$ I(k)=\int_k^{k^2}\frac{1}{x}\log\frac{x-1}{32}dx. $$ Then $$ I'(k)=\frac{2}{k}\log\frac{k^2-1}{32}-\frac{1}{k}\log\frac{k-1}{32}=\frac{1}{k}\log\frac{(k+1)^2(k-1)}{32}. $$ We have $I'(k)<0=I'(3)$ for $1<k<3$, $I'(k)>0=I'(3)$ for $k>3$, therefore $I(3)$ is the minimal value of $I(k)$ for $k>1$.

share|cite|improve this answer

$J=\int_k^{k^2}dx \frac{1}{x}\ln\frac{x-1}{32}=-\operatorname{dilog}(k)-\ln(\frac{k}{32}-\frac{1}{32})\ln(k)+\operatorname{dilog}(k^2)+\ln(\frac{k^2}{32}-\frac{1}{32})\ln(k^2)$ where: $$\operatorname{dilog}(x)=\int_1^x \, dt \ln(\frac{t}{1-t})$$ The derivative respect to $k$ is: $$\frac{\partial J}{\partial k}=-\frac{1}{k}[5 \ln(2)+\ln(k-1)-2\ln((k-1)(k+1))$$ putting: $$\frac{\partial J}{\partial k}=0$$ the result is $k=3$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.