Is limn→∞∫0π2x1n+sin⁡xtan⁡x+x1ndx=∫0π21+sin⁡xtan⁡x+1dx=1.62..?Note: The LHS of this identity it's seems converges for every positive integer n≥1 over the range (0,π2)

Noting0≤sin⁡x≤x,tan⁡x≥x,x∈[0,π2]one has0≤x1n+sin⁡xtan⁡x+x1n≤x1n+xx+x1n=1which implieslimn→∞∫0π2x1n+sin⁡xtan⁡x+x1ndx=∫0π2limn→∞x1n+sin⁡xtan⁡x+x1ndx=∫0π21+sin⁡xtan⁡x+1dxby using the DCT. Now under tan⁡(x2)→x , one has∫0π21+sin⁡xtan⁡x+1dx=2∫01(2x+1)(x2−1)(x2+1)(x2−2x−1)dxwhich can be handled by partial fractions easily.

"Is limn→∞∫0π2x1n+sin⁡xtan⁡x+x1ndx=∫0π21+sin⁡xtan⁡x+1dx=1.62..?Note: The LHS of this identity it's..." was answered by students like you

gabolzm6d

Answered question

2022-04-14

lim_{n \to \infty} \int_{0}^{\frac{π}{2}} \frac{x^{\frac{1}{n}} + \sin x}{\tan x + x^{\frac{1}{n}}} d x = \int_{0}^{\frac{π}{2}} \frac{1 + \sin x}{\tan x + 1} d x = 1.62 . .

?
Note: The LHS of this identity it's seems converges for every positive integer

n \geq 1

over the range

(0, \frac{π}{2})

Answer & Explanation

firenzesunzc65

Beginner2022-04-15Added 16 answers

Noting

0 \leq \sin x \leq x, \tan x \geq x, x \in [0, \frac{π}{2}]

one has

0 \leq \frac{x^{\frac{1}{n}} + \sin x}{\tan x + x^{\frac{1}{n}}} \leq \frac{x^{\frac{1}{n}} + x}{x + x^{\frac{1}{n}}} = 1

which implies

lim_{n \to \infty} \int_{0}^{\frac{π}{2}} \frac{x^{\frac{1}{n}} + \sin x}{\tan x + x^{\frac{1}{n}}} d x = \int_{0}^{\frac{π}{2}} lim_{n \to \infty} \frac{x^{\frac{1}{n}} + \sin x}{\tan x + x^{\frac{1}{n}}} d x = \int_{0}^{\frac{π}{2}} \frac{1 + \sin x}{\tan x + 1} d x

by using the DCT. Now under

\tan (\frac{x}{2}) \to x

, one has

\int_{0}^{\frac{π}{2}} \frac{1 + \sin x}{\tan x + 1} d x = 2 \int_{0}^{1} \frac{(2 x + 1) (x^{2} - 1)}{(x^{2} + 1) (x^{2} - 2 x - 1)} d x

which can be handled by partial fractions easily.

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