Which $F$ & $G$ satisfy $\int_0^{\pi}F(\sin (x))\cos (x)\, dx=0$ &$\int_{-\pi/2}^{\pi/2}\sin (x)\,G(\cos (x))\,dx=0$?

Question

Which functions $F$ & $G$ satisfy:- $\int_0^{\pi}F(\sin (x))\cos(x)\,dx=0$ and $\int_{-\pi/2}^{\pi/2}\sin(x)\,G(\cos(x))\,dx=0?$

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Equivalent relationships are given in The Universal Encyclopedia of Mathematics(1976 edition) page 597. But the specifics of the functions $F$ and $G$ are not given.

The rules DO NOT apply for various functions such as:- $F(\sin(x))= x\,{\sin(x)}$, $F(\sin(x))=\exp(x)\,{\sin(x)}$ , $F(\sin(x))=\frac{\text{d}(\sin (x))}{\text{d} x}{\cos(x)}.$

It seems to me that the rules DO apply only for "pure" functions of $\sin (x)$ (for $F$) and $\cos (x)$ (for $G$) respectively.

"Pure" in this sense means, for example, that the following pattern is valid:- $F(\sin(x)) = m\sin^n(x)$ for real numbers: $m,n$.

Whereas functions $F(\sin(x))$ which contain some (any?) functions of $x$ other than $\sin(x)$ may be designated "Impure". Also included in the impure category would be differentials and integrals of $\sin(x)$ with respect to $x$.

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Question

Is there a succinct way of identifying the two domains "pure" and "impure".?

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Update 1

The question arose from overlooking the strict mathematical definition of "function of" . The cited Universal Encyclopedia of Mathematics, p227-228 presents: "In the strict mathematical interpretation $y$ is called a function of $x$ if, corresponding to each permissible value of $x$, a ### value of $y$ can be calculated or observed". Perhaps the words "single, definite" should be inserted at the position marked by ###?

According to this definition, $|x|$ is a function of $x^2$ but $x$ is not a function of $x^2$.

I also learned that $F(x) = \int f(x) \space\text{d}x"$ does not mean that $F(x)$ is a function of $f(x)$ in the strict, mathematical sense. Likewise for functions related by differentiation.

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Update 2

The above hypotheiszed categories:- "pure functions" and "impure functions" are represented by the formal categories:- "injective functions" and "non-injective functions" as described here:- injective functions.

For the rules in the title to be true the functions $F$ and $G$ must be injective functions. For example the function $$y_1 = F_1(x) =\arcsin(\sin(x))$$ is not injective because a single value of $\sin(x)$ maps to multiple values of $y_1$. An injective function can be derived by setting an appropriately restricted range of $x$, as in the function $$y_2 = F_2(x) =\arcsin(\sin(x)) \space{} (-\pi/2 <= x <= \pi/2).$$

Answer 1

The first problem is $[\widetilde{F}(\sin x)]_0^\pi=0$ for your favourite antiderivative $\widetilde{F}$ of $F$, so any continuous $F$ integrable on $[0,\,1]$ will work. Similarly, your second problem is $[\widetilde{G}(\cos x)]_{-\pi/2}^{\pi/2}=0$ for an antiderivative $\widetilde{G}$ of $G$, so you just need $G$ to be continuous and integrable on $[0,\,1]$.

Answer 2

Firstly, we have: $$\int_0^\pi F(\sin x)\cos (x)dx=0$$ if we let $u=\sin(x)$ then $dx=\frac{du}{\cos(x)}$ and the integral can also be split into: $$\int_0^\pi=\int_0^{\pi/2}+\int_{\pi/2}^\pi$$ so the problem can be simplified as follows: $$\int_0^1F(u)du+\int_1^0F(u)du=0$$ since the integral on the right is just the negative of the integral on the left, this is true for all $F(u)$ which are continuous and integratable for $u\in[0,1]$.

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Secondly, we have: $$\int_{-\pi/2}^{\pi/2}G(\cos x)\sin(x)dx=0$$ using the same method as above you can prove once again that this is true for any $G(u)$ for $u\in[0,1]$

Answer 3

It's not about pure vs impure... the function $x\sin x$ is simply not function of $\sin x$, and similarly for your other examples. Take $x = {2\pi \over 3}$ and $x = {\pi \over 3}$ for example. They both have $\sin x = {\sqrt{3} \over 2}$ but different values of $x\sin x$.

The function $\sin x\,G(\cos x)$ is an odd function, so basically in any situation where $\int_{-{\pi \over 2}}^{\pi \over 2}\sin x\,G(\cos x)\,dx$ is defined this integral will be zero. This can be seen for example by changing variables from $x$ to $-x$; when one does this the integral is simply multiplied by $-1$. The only number equal to its negative is zero, so the integral is zero.

The same idea works for the first integral, since the integrand $I(x) = F(\sin x) \cos x$ satisfies $I(\pi - x) = -I(x)$ and therefore the change of variables from $x$ to $\pi - x$ also multiplies the integral by $-1$. Hence the first integral is also zero.