show-that-0-pi-2-sin-nx-sin-x-dx-pi-2-n-is-posative-odd-number-

Question Number 83064 by M±th+et£s last updated on 27/Feb/20

s h o w t h a t

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

n i s p o s a t i v e o d d n u m b e r

Answered by mind is power last updated on 27/Feb/20

\int_{0}^{\frac{π}{2}} \frac{s i n (2 k x + x)}{s i n (x)} = \int_{0}^{\frac{π}{2}} {\frac{s i n (2 k x) c o s (x)}{s i n (x)} + c o s (2 k x)} d x

k = 0 e v i d e n t e

\underset{k = 0}{\sum^{n - 1}} e^{i k x} = \frac{1 - (e^{i n x})}{1 - e^{i x}} = \frac{e^{i (\frac{n - 1}{2}) x} s i n (\frac{n}{2} x)}{s i n (\frac{x}{2})}

\Rightarrow \underset{k = 0}{\sum^{n - 1}} e^{i 2 k x} = \frac{e^{i (n - 1) x} s i n (n x)}{s i n (x)}

\Rightarrow \underset{k = 0}{\sum^{n - 1}} e^{i (2 k - n + 1) x} = \frac{s i n (n x)}{s i n (x)}

n = 2 m + 1

\Rightarrow \int_{0}^{\frac{π}{2}} \underset{k = 0}{\sum^{2 m}} e^{i (2 k - 2 m) x} d x = \int_{0}^{\frac{π}{2}} \frac{s i n ((2 m + 1) x)}{s i n (x)} d x

m > 0 \Rightarrow

\int_{0}^{\frac{π}{2}} \underset{k = 0}{\sum^{m - 1}} e^{i (2 k - 2 m) x} d x + \int_{0}^{\frac{π}{2}} e^{i . (2 m - 2 m) x} d x + \int_{0}^{\frac{π}{2}} \underset{k = m + 1}{\sum^{2 m}} e^{i (2 k - 2 m) x} d x

= \underset{k = 0}{\sum^{m - 1}} \int_{0}^{\frac{π}{2}} e^{i (2 k - 2 m) x} d x

= \underset{k = 0}{\sum^{m - 1}} {[\frac{e^{i (2 k - 2 m) x}}{i (2 k - 2 m)}]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} d x + \underset{m + 1}{\sum^{2 m}} {[\frac{e^{i (2 k - 2 m) x}}{2 k - 2 m}]}_{0}^{\frac{π}{2}}

= \underset{k = 0}{\sum^{m - 1}} \frac{{(- 1)}^{k - m} - 1}{i 2 (k - m)} + \frac{π}{2} + \underset{m + 1}{\sum^{2 m}} [\frac{{(- 1)}^{k - m} - 1}{i 2 (k - m)}]

\underset{m + 1}{\sum^{2 m}} [\frac{{(- 1)}^{k - m} - 1}{2 i (k - m)}] = \underset{0}{\sum^{m - 1}} [\frac{{(- 1)}^{2 m - k - m} - 1}{2 i (2 m - k - m)}]

= \underset{k = 0}{\sum^{m - 1}} [\frac{{(- 1)}^{k - m} - 1}{2 i (m - k)}] = - \underset{k = 0}{\sum^{m - 1}} [\frac{{(- 1)}^{k - m} - 1}{2 i (k - m)}]

\Rightarrow \underset{m + 1}{\sum^{2 m}} [\frac{{(- 1)}^{k - m} - 1}{2 i (k - m)}] + \underset{k = 0}{\sum^{m - 1}} [\frac{{(- 1)}^{k - m} - 1}{2 i (k - m)}] = 0

\Rightarrow \underset{k = 0}{\sum^{m - 1}} \int_{0}^{\frac{π}{2}} e^{i 2 (k - m) x} d x = \frac{π}{2} = \int_{0}^{\frac{π}{2}} \frac{s i n (n x)}{s i n (x)} d x

Commented by M±th+et£s last updated on 27/Feb/20

g o d b l e s s y o u

Commented by mind is power last updated on 27/Feb/20

w i t h e p l e a s u r g o d b l e s s e v r e y o n e

Answered by TANMAY PANACEA last updated on 28/Feb/20

I_{n} - I_{n - 2} = \int_{0}^{\frac{π}{2}} \frac{s i n n x - s i n (n - 2) x}{s i n x} d x

= \int_{0}^{\frac{π}{2}} \frac{2 c o s (n - 1) x s i n x}{s i n x} d x

I_{n} - I_{n - 2} = 2 ∣ \frac{s i n (n - 1) x}{n - 1} ∣_{0}^{\frac{π}{2}} = \frac{2}{n - 1} s i n (n - 1) \frac{π}{2} = 0

★ w h e n n = o d d t h e n n - 1 = e v e n

s o s i n (n - 1) \frac{π}{2} = s i n (2 k \times \frac{π}{2}) = 0 ★

I_{n} = I_{n - 2}

I_{3} = I_{1}

I_{5} = I_{3}

s o I_{1} = I_{3} = I_{5} = I_{7} \dots .

I_{1} = \int_{0}^{\frac{π}{2}} \frac{s i n x}{s i n x} d x =∣ x ∣_{0}^{\frac{π}{2}} = \frac{π}{2}

s o I_{n} = \frac{π}{2}

Commented by M±th+et£s last updated on 28/Feb/20

g o d b l e s s y o u s i r

Commented by TANMAY PANACEA last updated on 28/Feb/20

b l e s s i n g s h o w e r t o a l l