# Find the minimum value of this integral.

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?

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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

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 $k>1$.
$J=\int_k^{k^2}dx \frac{1}{x}\ln\frac{x-1}{32}=-\operatorname{dilog}(k)-\ln(\frac{k}{32}-\frac{1}{32})\ln(k)+\operatorname{dilog}(k^2)+\ln(\frac{k^2}{32}-\frac{1}{32})\ln(k^2)$ 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 $k=3$