# Prove that $\int_{0}^{1}f(x)\arctan x dx=\frac{π}{8}\int_{0}^{1}f(x)dx$

For $$f:[0,1]\rightarrow\mathbb{R}$$ a continuous function with the property that $$2f(\frac{1-x}{1+x})=(1+x)^2$$. Prove that $$\int_{0}^{1}f(x)\arctan xdx=\frac{π}{8}\int_{0}^{1}f(x)$$.

However, I obtained that $$\int_{0}^{1}f(x)\arctan xdx=\int_{0}^{1} 2f\left(\frac{1-t}{1+t}\right)\frac{1}{(1+t)^2}\arctan\left(\frac{1-t}{1+t}\right)dt=\frac{π}{4}-\int_{0}^{1}\arctan tdt$$ Which is $$\frac{\ln 2}{2}$$. However $$\int_{0}^{1}f(x)dx=1$$ after similar changes in variable. The results don't match, in conclusion.

Edit: I copied the exercise correctly. I post a photo of it, sorry for being in Romanian.

• I believe you copied the exercise wrong. You missed out an $f(x)$. – Number Mar 6 at 11:42
• @SeptimiuCristian Title and author of this romanian book? – Robert Z Mar 6 at 12:11
• Indeed, it would be a good idea to inform the author. Although I see there the year is $1999$ so it might be kinda late. – Number Mar 6 at 12:12
• @RobertZ It's from the supliment of ,,Gazeta Matematica", no. 1, 2019, – Septimiu Cristian Mar 6 at 12:17
• @Zacky I think the same if indeed the mistake is in the original exercise. – Septimiu Cristian Mar 6 at 12:19

Assuming that you meant:$$2f\left(\frac{1-x}{1+x}\right)=\color{orange}{f(x)}(1+x)^2$$ We can start with the given integral and let $$x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$$ $$I=\int_{0}^{1}f(x)\arctan xdx=\int_0^1 \color{blue}{f\left(\frac{1-t}{1+t}\right)}\color{red}{\arctan \left(\frac{1-t}{1+t}\right)}\frac{\color{blue}{2}}{(1+t)^2}dt$$ $$=\int_0^1 \color{blue}{f(t)}\left(\color{red}{\frac{\pi}{4}-\arctan t}\right)dt\overset{t=x}=\frac{\pi}{4}\int_0^1 f(x)dx-I$$ $$\Rightarrow 2I=\frac{\pi}{4}\int_0^1 f(x)dx \Rightarrow I=\frac{\pi}{8}\int_0^1 f(x)dx$$
Your approach is correct but, as Zacky pointed out, a factor $$f(x)$$ is missing in the stated property of $$f$$.
Let $$x=\frac{1-t}{1+t}$$, then $$dx=\frac{-2dt}{(1+t^2)}$$, $$\arctan(x)=\arctan(1)-\arctan(t),$$ and, by using the given property $$2f(x)=f(t)(1+t)^2$$ (with $$f$$), we get $$\int_{0}^{1}f(x)\arctan(x)\,dx=\int_{0}^{1}f(t)(\arctan(1)-\arctan(t))dt\\ = \frac{\pi}{4}\int_{0}^{1}f(t)\,dt-\int_{0}^{1}f(t)\arctan(t))\,dt$$ which implies that $$\int_{0}^{1}f(x)\arctan(x)\,dx=\frac{\pi}{8}\int_{0}^{1}f(t)\,dt.$$
• How did you got this:$$\int_{0}^{1}f(x)\arctan(x)\,dx=\int_{0}^{1}f(t)(\arctan(1)-\arctan(t))dt?$$With: $2f\left(\frac{1-t}{1+t}\right)=(1+t)^2$ – Number Mar 6 at 11:51
• With that substitution, one gets: $$\int_{0}^{1}f(x)\arctan xdx=\int_0^1 2{f\left(\frac{1-t}{1+t}\right)}{\arctan \left(\frac{1-t}{1+t}\right)}\frac{{1}}{(1+t)^2}dt$$ Or I'm wrong? How there is still left with an $f(t)$ after you replace $2f\left(\frac{1-t}{1+t}\right)$ with $(1+t)^2$? – Number Mar 6 at 11:55
• I obtain that $\int_{0}^{1}f(x)\arctan(x)\,dx is \int_{0}^{1}f(\frac{1-t}{1+t})(\arctan(1)-\arctan(t))dt\\$. – Septimiu Cristian Mar 6 at 11:56