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

I'm interested in proving the following integral inequality:

$$\frac1{20}\le \int_{1}^{\sqrt 2} \frac{\ln x}{\ln^2x+1} dx$$

According to W|A the result of this integral isn't pretty nice, and involves the exponential integral.

share|cite|improve this question
up vote 6 down vote accepted

Let $f(x) = \dfrac{\ln(x)}{\ln^2(x) + 1}$. $f(x)$ is concave in $[1,\sqrt{2}]$.

enter image description here Hence, the area of $f(x)$ from $1$ to $\sqrt{2}$ i.e. the area under the blue curve is greater than the area of the triangle between the two points at which the blue curve and the red curve intersect i.e. \begin{align} \int_1^{\sqrt{2}} f(x) dx & \geq \dfrac12 \times (\sqrt{2} - 1) \times (f(\sqrt{2}) - f(1)) = \dfrac{\sqrt{2}-1}2 \times \dfrac{\ln(\sqrt{2})}{\ln^2(\sqrt{2})+1}\\ & = \dfrac{(\sqrt{2}-1)\ln(2)}{\ln^2(2)+4} \approx 0.06408 > \dfrac1{20} \end{align}

share|cite|improve this answer
That's a nice solution! Thanks! – user 1618033 Jun 22 '12 at 7:40

$$\begin{array}{c l} \int_1^{\sqrt2} \frac{\log x}{1+(\log x)^2}dx & =\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}e^udu \\ & \ge\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}du \\ & = \left[\frac{1}{2}\log(1+u^2)\right]_0^{\frac{1}{2}\log 2} \\ & =\frac{1}{2}\log\left(1+\frac{(\log 2)^2}{4}\right) \\ & \approx 0.056714899 \\ & > 0.05=\frac{1}{20}. \end{array}$$

Edit. We can make it tighter with $e^u\ge1+u$:

$$\begin{array}{c l} \int_1^{\sqrt2} \frac{\log x}{1+(\log x)^2}dx & =\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}e^udu \\ & \ge\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}(1+u)du \\ & = \int_0^{\frac{1}{2}\log2}\left(1+\frac{u}{1+u^2}-\frac{1}{1+u^2}\right)du \\ & = \left[u+\frac{1}{2}\log(1+u^2)-\arctan u\right]_0^{\frac{1}{2}\log 2} \\ & =\frac{1}{2}\log\left(2+\frac{(\log 2)^2}{2}\right)-\arctan\left(\frac{1}{2}\log 2\right) \\ & \approx 0.06\color{Blue}{96694}. \end{array}$$

share|cite|improve this answer
+1. This was the first one I tried and was about to write this up. But then the triangle argument gave a better lower bound and wrote it up. – user17762 Jun 22 '12 at 8:01
@anon: nice! Thanks! Being inspired by your solution i think it also works to write $\int_1^{\sqrt2} \frac{\log x}{x(1+(\log x)^2)}dx \le \int_1^{\sqrt2} \frac{\log x}{1+(\log x)^2}dx$ if i'm not wrong. – user 1618033 Jun 22 '12 at 8:03
@Marvis: I just now noticed that if we use $e^u\ge1+u$ instead of merely $e^u\ge1$, we get an even better bound than the triangle argument. I call dibs! – anon Jun 22 '12 at 8:04
@Marvis: probably in such cases as a part of the art of problem solving, we should look after triangle argument firstly. :-) – user 1618033 Jun 22 '12 at 8:09

It's enough to compare logarithm with some linear function. We have: $\ln x \ge \frac{\ln 2}{2( \sqrt{2} - 1)} (x-1), \;\; x\in [1, \sqrt{2}]$ and $\ln x \le x-1, \; \; x \ge 1$ so:

$$\int_1^{\sqrt{2}} \frac{\ln x \; \mbox d x}{\ln^2 x + 1} \ge \int_1^{\sqrt{2}} \frac{\frac{\ln 2}{2( \sqrt{2} - 1)} (x-1)}{1+(x-1)^2} \, dx = \frac{1+ \sqrt{2}}{4} \ln (8- 4 \sqrt{2}) \approx 0.066 > 0.05$$

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.