I-n-0-pi-2-sin-2-nt-sin-t-dt-Find-lim-n-2I-n-lnn-

Question Number 160395 by qaz last updated on 29/Nov/21

I_{n} = \int_{0}^{π / 2} \frac{\sin^{2} (nt)}{\sin t} dt

Find :: \lim_{n \to \infty} ({2 I}_{n} - lnn) = ?

Answered by Kamel last updated on 29/Nov/21

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

I_{n + 1} - I_{n} = 4 \int_{0}^{\frac{π}{2}} \frac{s i n (\frac{2 n + 1}{2} x) c o s (\frac{x}{2}) s i n (\frac{x}{2}) c o s (\frac{2 n + 1}{2})}{s i n (x)} d x

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

I_{n + 1} = \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots + \frac{1}{2 n + 1} + 1 = H_{2 n + 1} - \frac{1}{2} H_{n}

∴ L = \underset{n \to + \infty}{l i m} (2 I_{n} - L n (n)) = \underset{n \to + \infty}{l i m} (2 H_{2 n + 1} - H_{n} - L n (n + 1))

= \underset{n \to + \infty}{l i m} (2 (H_{2 n + 1} - L n (2 n + 1)) + 2 L n (2 n + 1) - H_{n} - L n (n + 1))

= \underset{n \to + \infty}{l i m} (2 (H_{2 n + 1} - L n (2 n + 1)) + 2 L n (\frac{2 n + 1}{n + 1}) - (H_{n + 1} - L n (n + 1)))

= γ + 2 L n (2)

∴ \underset{n \to + \infty}{l i m} 2 \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} (n x)}{s i n (x)} d x - L n (n) = 2 L n (2) + γ

W h e r e : γ = \underset{n \to + \infty}{l i m} (H_{n} - L n (n)) d e n o t e E u l e r - M a s c h e r o n i c o n s t a n t .

A n d : H_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1} + \frac{1}{n} t h e n^{t h} h a r m o n i c n u m b e r .

B E N A I C H A K A M E L

Commented by qaz last updated on 30/Nov/21

thanks a lot . very nice solution .

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