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Definite Integration: Question

JEE Maths question with a full step-by-step solution.

Question
In=∫n/2(n+1)/2sin⁡(πsin⁡2πx)(2)xdx,n∈II_n = \int_{n/2}^{\left(n+1\right)/2}\frac{\sin\left(\pi\sin^2\pi x\right)}{\left(\sqrt2\right)^x}dx , \qquad n \in I
AInIn+4=2\dfrac{I_n}{I_{n+4}} = 2correct
BInIn+4=12\dfrac{I_n}{I_{n+4}} = \dfrac{1}{\sqrt2}
C∑n=0∞I8nI0=43\dfrac{\displaystyle\sum_{n=0}^{\infty}I_{8n}}{I_0} = \dfrac43correct
D∑n=0∞InI0=2\dfrac{\displaystyle\sum_{n=0}^{\infty}I_n}{I_0} = 2
Solution
Step 1: Put x=k2+tx = \dfrac k2+t in In+kI_{n+k}, kk an integer; as xx runs over [n+k2,n+k+12]\left[\tfrac{n+k}2,\tfrac{n+k+1}2\right], tt runs over [n2,n+12]\left[\tfrac n2,\tfrac{n+1}2\right]:
In+k=∫n/2(n+1)/2sin⁡(πsin⁡2π(k2+t))(2)k/2+t dtI_{n+k} = \int_{n/2}^{\left(n+1\right)/2} \frac{\sin\left(\pi\sin^2\pi\left(\tfrac k2+t\right)\right)}{\left(\sqrt2\right)^{k/2+t}}\,dt
 Step 2: If kk is even, sin⁡π(k2+t)=±sin⁡πt\sin\pi\left(\tfrac k2+t\right) = \pm\sin\pi t, so sin⁡2π(k2+t)=sin⁡2πt\sin^2\pi\left(\tfrac k2+t\right) = \sin^2\pi t and the numerator is identical. If kk is odd, sin⁡π(k2+t)=±cos⁡πt\sin\pi\left(\tfrac k2+t\right) = \pm\cos\pi t, so sin⁡2→cos⁡2=1−sin⁡2πt\sin^2 \to \cos^2 = 1-\sin^2\pi t, and
sin⁡(π(1−sin⁡2πt))=sin⁡(π−πsin⁡2πt)=sin⁡(πsin⁡2πt)\sin\left(\pi\left(1-\sin^2\pi t\right)\right) = \sin\left(\pi-\pi\sin^2\pi t\right) = \sin\left(\pi\sin^2\pi t\right)
 the same again. So the numerator is invariant for **every** integer kk. Step 3:
(2)k/2+t=(2)k/2(2)t=2k/4(2)t\left(\sqrt2\right)^{k/2+t} = \left(\sqrt2\right)^{k/2}\left(\sqrt2\right)^{t} = 2^{k/4}\left(\sqrt2\right)^{t}
 Hence
 In+k=In2k/4 for every integer k\boxed{\ I_{n+k} = \frac{I_n}{2^{k/4}}\ }\qquad\text{for every integer }k
 Step 4: 0≤πsin⁡2πx≤π0 \le \pi\sin^2\pi x \le \pi, so sin⁡(πsin⁡2πx)≥0\sin\left(\pi\sin^2\pi x\right) \ge 0, and it is not identically zero on a half-unit interval. Hence every In>0I_n>0 and the ratios below are defined. Step 5: k=4k = 4 gives In+4=In21=In2I_{n+4} = \dfrac{I_n}{2^{1}} = \dfrac{I_n}{2}, so
InIn+4=2\frac{I_n}{I_{n+4}} = 2
 (1) holds; (2) fails. Step 6:
I8n=I028n/4=I04n,∑n=0∞I8n=I0∑n=0∞4−n=I0⋅11−14=43I0I_{8n} = \frac{I_0}{2^{8n/4}} = \frac{I_0}{4^{n}} ,\qquad \sum_{n=0}^{\infty}I_{8n} = I_0\sum_{n=0}^{\infty}4^{-n} = I_0\cdot\frac{1}{1-\tfrac14} = \frac43I_0
 (3) holds. Step 7:
∑n=0∞In=I0∑n=0∞2−n/4=I01−2−1/4=I01−0.840896=6.2852 I0≠2I0\sum_{n=0}^{\infty}I_n = I_0\sum_{n=0}^{\infty}2^{-n/4} = \frac{I_0}{1-2^{-1/4}} = \frac{I_0}{1-0.840896} = 6.2852\,I_0 \ne 2I_0
 (4) fails. Answer: (1) and (3)
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