After I was playing with the formulas $\cos(a\pm b)$ and [1], I wondered a different case from those calculations that I was doing in the context to study the formula of [1].
Question. Let $\operatorname{frac}(x)= \left\{ x \right\}$ the fractional part function. Is it possible calculate a closed-form for $$\int_0^7\sin\left(\frac{2\pi\cdot 3x}{5}\right)\sin\left(\frac{2\pi\cdot 15 \left\{ x \right\}}{8}\right)dx\,?\tag{I}$$ Then, please provide me a way to get it. Many thanks.
I presume that using Wolfram Alpha online calculator, and my knowledges (the fractional part is periodic) that I can to prove closed-forms for different integrals but more simple than previous. Wolfram Alpha online calculator (using standard time of computation), provide me the output
int sin(2 pi 3 x/5)sin(2 pi 15 frac(x)/8)dx, from x=0 to 7
Now as comparison look at from this different input, the closed-form that Wolfram Alpha provide me
int sin(2 pi 3 x/5)sin(2 pi 15 frac(x)/5)dx, from x=0 to 7
That I am asking this: is if we can get a closed-form for the integral (I), and how do it.
References:
[1] George Purdy, An Integral Equal to $\sigma(n)$, Problems and Solutions, Problem E 1850 [1966, 82], The American Mathematical MONTHLY, Vol. 74 N. 5, p. 594-595 (MAY, 1967).
Best Answer
This is more of a general roadmap.
First, one may show that, for continuous and integrable functions $f$ and $g$:
$$\int_0^N dx \, f(x) g(\{x\}) = \int_0^1 dx \, g(x) \sum_{k=0}^{N-1} f(x+k)$$
where $N \in \mathbb{N}$. For your integral, the sum boils down to
$$\sum_{k=0}^7 \sin{\left ( \frac{6 \pi}{5} (x+k) \right )} = A \cos{\left ( \frac{6 \pi}{5} x \right )} + B \sin{\left ( \frac{6 \pi}{5} x \right )}$$
where $A$ and $B$ are, respectively, imaginary and real parts of the quantity $1/(e^{-i 2 \pi/5}+1)$. This then reduces to two integrals:
$$\int_0^1 dx \, \sin{\left ( \frac{15 \pi}{4} x \right )} \cos{\left ( \frac{6 \pi}{5} x \right )} $$
$$\int_0^1 dx \, \sin{\left ( \frac{15 \pi}{4} x \right )} \sin{\left ( \frac{6 \pi}{5} x \right )} $$
which I assume you can do.