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Question Number 83074 by ~blr237~ last updated on 27/Feb/20

$$findallfunctionsatisfying\mathrm{\forall }x\in \mathbb{R}\mathrm{\setminus }\{k\pi ,k\in \mathbb{Z}\}$$$$f\left(x\right)+{\int }_0^1{f}^2\left(x\right)dx=\frac{x}{sin\left(\pi x\right)}$$

Commented by mr W last updated on 28/Feb/20

$${\int }_0^1{f}^2\left(x\right)dx=k=constant$$$$f\left(x\right)=\frac{x}{sin\left(\pi x\right)}-k$$$$f^2\left(x\right)=\frac{x^2}{{sin}^2\left(\pi x\right)}-2k\frac{x}{\sin \pi x}+k^2$$$${\int }_0^1{f}^2\left(x\right)dx={\int }_0^1\frac{x^{2}dx}{{sin}^2\left(\pi x\right)}-2k{\int }_0^1\frac{xdx}{\sin \pi x}+k^2=k$$$$k^2-(1+2{\int }_0^1\frac{xdx}{\sin \pi x})k+{\int }_0^1\frac{x^{2}dx}{{sin}^2\left(\pi x\right)}=0$$$$k^2-(1+2A)k+B=0$$$$\Rightarrow k=\frac{1+2A\pm \sqrt{{(1+2A)}^2-4B}}{2}$$$$\Rightarrow f\left(x\right)=\frac{x}{sin\left(\pi x\right)}-\frac{1+2A\pm \sqrt{{(1+2A)}^2-4B}}{2}$$$$with$$$$A={\int }_0^1\frac{xdx}{\sin \pi x}=\frac{1}{{\pi }^2}{\int }_0^{\pi }\frac{tdt}{\sin t}$$$$B={\int }_0^1\frac{x^{2}dx}{{sin}^2\left(\pi x\right)}=\frac{1}{{\pi }^3}{\int }_0^{\pi }\frac{t^{2}dt}{{\sin }^{2}t}$$$$butAandBmaynotexist,sothe$$$$questionmaybewrong.$$

Commented by ~blr237~ last updated on 28/Feb/20

$$nicesir!theboth{don}^{\prime }texist:{it}^{\prime }sfeelassometimes{it}^{\prime }snoteasytoprovethatthesolutionsetofan$$$$equationisempty$$

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