number of distinct solution of Integral equation
Total number of distinct $0\leq x\leq 1$ for which $\displaystyle \int_{0}^{x}\frac{t^2}{1+t^4}dt = 2x-1$
$\bf{My\; Try::}$ Given $\displaystyle \underbrace{\int_{0}^{x}\frac{t^2}{1+t^4}dt}_{\geq 0} = 2x-1\;,$ Where $0\leq x\leq 1$
So we get $\displaystyle x\geq \frac{1}{2}\;\;\;,$ Now $\displaystyle \frac{d}{dx}\left\{\int_{0}^{x}\frac{t^2}{1+t^4}dt\right\} =\frac{d}{dx}(2x-1)$
So $$\frac{x^2}{1+x^4}=2\Rightarrow \underbrace{x^2+\frac{1}{x^2}}_{\geq 2} = \frac{1}{2}$$
So no real values of $x$
Is my solution is Right, If not then how can i solve it some short way, Thanks
$\endgroup$ 52 Answers
$\begingroup$$$ x\mapsto \frac{x^2}{1+x^4} $$ is increasing on $[0,1]$, hence $f(x)=\int_{0}^{x}\frac{t^2}{1+t^4}\,dt $ is non-negative and convex on $[0,1]$ and $f'(x)\leq f'(1)=\frac{1}{2}$ holds. It follows that $g(x)=f(x)-2x+1$ is decreasing on $[0,1]$. Since $g(0)=1>0$ and $$ g(1) = \frac{\pi}{4\sqrt{2}}-\frac{\log(1+\sqrt{2})}{2\sqrt{2}}-1 <0$$ it follows that $g(x)=0$ has only one solution in $[0,1]$.
$\endgroup$ 1 $\begingroup$Just to find out the only solution whose existence has been given by Jack.
Integrating $$\displaystyle \int_{0}^{x}\frac{t^2}{1+t^4}dt=\frac{1}{4\sqrt 2}\left(\log \left(\frac{x^2-\sqrt 2 x+1}{x^2+\sqrt 2 x+1}\right)+2\arctan(1+\sqrt2)-2\arctan(1-\sqrt2)\right) $$ Now graphing $f(x)=\displaystyle \int_{0}^{x}\frac{t^2}{1+t^4}dt$ without the denominator $4\sqrt2$ and $g(x)=4\sqrt2(2x-1)$ we get
In which the only point of intersection is approximately $(x,y)=(0.523,0.262)$ hence $$\color{red}{x\approx 0.523}$$
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