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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.enter image description here

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  • $\begingroup$ I believe you copied the exercise wrong. You missed out an $f(x)$. $\endgroup$ – Number Mar 6 at 11:42
  • $\begingroup$ @SeptimiuCristian Title and author of this romanian book? $\endgroup$ – Robert Z Mar 6 at 12:11
  • $\begingroup$ 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. $\endgroup$ – Number Mar 6 at 12:12
  • $\begingroup$ @RobertZ It's from the supliment of ,,Gazeta Matematica", no. 1, 2019, $\endgroup$ – Septimiu Cristian Mar 6 at 12:17
  • $\begingroup$ @Zacky I think the same if indeed the mistake is in the original exercise. $\endgroup$ – Septimiu Cristian Mar 6 at 12:19
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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 $x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$ $$I=\int_{0}^{1}f(x)\arctan xdx=\int_0^1 \color{blue}{f\left(\frac{1-t}{1+t}\right)}\color{red}{\arctan \left(\frac{1-t}{1+t}\right)}\frac{\color{blue}{2}}{(1+t)^2}dt$$ $$=\int_0^1 \color{blue}{f(t)}\left(\color{red}{\frac{\pi}{4}-\arctan t}\right)dt\overset{t=x}=\frac{\pi}{4}\int_0^1 f(x)dx-I$$ $$\Rightarrow 2I=\frac{\pi}{4}\int_0^1 f(x)dx \Rightarrow I=\frac{\pi}{8}\int_0^1 f(x)dx$$

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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 $$\int_{0}^{1}f(x)\arctan(x)\,dx=\frac{\pi}{8}\int_{0}^{1}f(t)\,dt.$$

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  • $\begingroup$ How did you got this:$$ \int_{0}^{1}f(x)\arctan(x)\,dx=\int_{0}^{1}f(t)(\arctan(1)-\arctan(t))dt?$$With: $2f\left(\frac{1-t}{1+t}\right)=(1+t)^2$ $\endgroup$ – Number Mar 6 at 11:51
  • $\begingroup$ 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 $(1+t)^2$? $\endgroup$ – Number Mar 6 at 11:55
  • $\begingroup$ I obtain that $\int_{0}^{1}f(x)\arctan(x)\,dx is \int_{0}^{1}f(\frac{1-t}{1+t})(\arctan(1)-\arctan(t))dt\\$. $\endgroup$ – Septimiu Cristian Mar 6 at 11:56
  • $\begingroup$ However, considering Zacky's suggestion, your solution is correct. $\endgroup$ – Septimiu Cristian Mar 6 at 11:58
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    $\begingroup$ Sorry, I misread the statement. Yes a correction is needed! $\endgroup$ – Robert Z Mar 6 at 12:02

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