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

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.

-

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

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}

-
That's a nice solution! Thanks! –  Chris's sis the artist 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 $e^u\ge1+u$:

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

-
+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. –  Chris's sis the artist 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. :-) –  Chris's sis the artist 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:

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

-