Finding the value of given integral.

Question

It was asked to find the correct option(s) for the given integral: $$I_n = \displaystyle\int_{\frac{n}{2}}^{\frac{n+1}{2}}\dfrac{\sin{(\pi(\sin^2{\pi x}}))}{(\sqrt2)^x} \, dx$$

(a)$\dfrac{I_n}{I_{n+4}}=2$

(b)$\dfrac{I_n}{I_{n+4}}=\dfrac{1}{\sqrt2}$

(c)$\dfrac{\displaystyle\sum_{n=0}^\infty I_{8n}}{I_0}=\dfrac{4}{3}$

(d)$ \dfrac{\displaystyle\sum_{n=0}^\infty I_{n}}{I_0}=2$

I tried using King's property to solve this one, but it is not working here, instead it is making it more complicated.

Any help is appreciated :)

Will you plz bracket it properly initially I though $\pi \sin^2\pi$ is a constant, which is not.Hint: Try option chasing lol!!! It works. Also 1st option can be checked by plugging n=0 also for the second option, Note in this problem first two options are invariant in values. — , Jan 10 '23 at 17:28
Try computing $I_{n+4}$ and using a substitution that will make $I_{n}$ appears — Raul Fernandes Horta, Jan 10 '23 at 17:30
"find the correct options", as in select all that are true? Or do you only expect one of these answers to be correct? — user170231, Jan 10 '23 at 17:46

score 4 · Answer 1 · answered Jan 10 '23 at 21:24

First, notice that for any even integer $2k$

$$I_{n+2k} = \int_{\frac{n}{2}+k}^{\frac{n+1}{2}+k}\frac{\sin\left[\pi \sin^2(\pi x)\right]}{\left(\sqrt{2}\right)^x}dx = \int_{\frac{n}{2}}^{\frac{n+1}{2}}\frac{\sin\left[\pi \sin^2(\pi x + \pi k)\right]}{\left(\sqrt{2}\right)^{x+k}}dx = \frac{1}{\left(\sqrt{2}\right)^k}I_n$$

Selecting $k=2$ gives us

$$\frac{I_n}{I_{n+4}} = \frac{I_n}{\frac{1}{\left(\sqrt{2}\right)^2}I_n} = 2$$

Selecting $k=4n$ gives us

$$\frac{\sum\limits_{n=0}^{\infty}I_{8n}}{I_0} = \frac{\sum\limits_{n=0}^\infty \frac{1}{\left(\sqrt{2}\right)^{4n}}I_0}{I_0} = \sum_{n=0}^\infty \frac{1}{4^n} = \frac{4}{3}$$

Selecting $k=2n$ gives us

$$\frac{\sum\limits_{n=0}^\infty I_n}{I_0} > \frac{\sum\limits_{n=0}^\infty I_{4n}}{I_0} = \frac{\sum\limits_{n=0}^\infty \frac{1}{\left(\sqrt{2}\right)^{2n}}I_0}{I_0} = \sum_{n=0}^\infty \frac{1}{2^n} = 2$$

because

$$\pi \sin^2(\pi x) \in [0,\pi] \implies I_n > 0$$

Therefore, the only correct choices are options $(a)$ and $(c)$

Finding the value of given integral.

1 Answers1