find-all-function-satisfying-x-R-kpi-k-Z-f-x-0-1-f-2-x-dx-x-sin-pix-

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

f i n d a l l f u n c t i o n s a t i s f y i n g \forall x \in R ∖ {k π, k \in Z}

f (x) + \int_{0}^{1} f^{2} (x) d x = \frac{x}{s i n (π x)}

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

\int_{0}^{1} f^{2} (x) d x = k = c o n s t a n t

f (x) = \frac{x}{s i n (π x)} - k

f^{2} (x) = \frac{x^{2}}{{s i n}^{2} (π x)} - 2 k \frac{x}{\sin π x} + k^{2}

\int_{0}^{1} f^{2} (x) d x = \int_{0}^{1} \frac{x^{2} d x}{{s i n}^{2} (π x)} - 2 k \int_{0}^{1} \frac{x d x}{\sin π x} + k^{2} = k

k^{2} - (1 + 2 \int_{0}^{1} \frac{x d x}{\sin π x}) k + \int_{0}^{1} \frac{x^{2} d x}{{s i n}^{2} (π x)} = 0

k^{2} - (1 + 2 A) k + B = 0

\Rightarrow k = \frac{1 + 2 A \pm \sqrt{{(1 + 2 A)}^{2} - 4 B}}{2}

\Rightarrow f (x) = \frac{x}{s i n (π x)} - \frac{1 + 2 A \pm \sqrt{{(1 + 2 A)}^{2} - 4 B}}{2}

w i t h

A = \int_{0}^{1} \frac{x d x}{\sin π x} = \frac{1}{π^{2}} \int_{0}^{π} \frac{t d t}{\sin t}

B = \int_{0}^{1} \frac{x^{2} d x}{{s i n}^{2} (π x)} = \frac{1}{π^{3}} \int_{0}^{π} \frac{t^{2} d t}{\sin^{2} t}

b u t A a n d B m a y n o t e x i s t, s o t h e

q u e s t i o n m a y b e w r o n g .

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

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e q u a t i o n i s e m p t y