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Question Number 59185 by maxmathsup by imad last updated on 09/May/19

c a l c u l a t e \int_{0}^{1} \frac{x}{s i n x} d x

l e t f (x) = s i n x \Rightarrow f (x) = f (0) + {x f}^{^{'}} (0) + \frac{x^{2}}{2} f^{(2)} (0) + \frac{x^{3}}{3!} f^{(3)} (0) + o (x^{4})

b u t f (0) = 0 f^{^{'}} (x) = c o s x \Rightarrow f^{^{'}} (0) = 1 f^{(2)} (x) = - s i n x \Rightarrow f^{(2)} (0) = 0

f^{(3)} (0) = - c o s (x) \Rightarrow f^{(3)} (0) = - 1 \Rightarrow s i n x = x - \frac{x^{3}}{6} + o (x^{4}) \Rightarrow

x - \frac{x^{3}}{3} ⩽ s i n x ⩽ x f o r x \in] 0, 1] \frac{1}{x} ⩽ \frac{1}{s i n x} ⩽ \frac{1}{x - \frac{x^{3}}{6}} \Rightarrow 1 ⩽ \frac{x}{s i n x} ⩽ \frac{1}{1 - \frac{x^{2}}{6}} \Rightarrow

1 ⩽ \int_{0}^{1} \frac{x}{s i n x} d x ⩽ \int_{0}^{1} \frac{d x}{1 - \frac{x^{2}}{6}}

\int_{0}^{1} \frac{d x}{1 - \frac{x^{2}}{6}} =_{x = \sqrt{6} t} \int_{0}^{\frac{1}{\sqrt{6}}} \frac{\sqrt{6} d t}{1 - t^{2}} = \frac{\sqrt{6}}{2} \int_{0}^{\frac{1}{\sqrt{6}}} (\frac{1}{1 - t} + \frac{1}{1 + t}) d t

= \frac{\sqrt{6}}{2} {[l n ∣ \frac{1 + t}{1 - t} ∣]}_{0}^{\frac{1}{\sqrt{6}}} = \frac{\sqrt{6}}{2} {l n (\frac{1 + \frac{1}{\sqrt{6}}}{1 - \frac{1}{\sqrt{6}}})} = \frac{\sqrt{6}}{2} l n (\frac{\sqrt{6} + 1}{\sqrt{6} - 1}) \Rightarrow

1 ⩽ \int_{0}^{1} \frac{x}{s i n x} d x ⩽ \frac{\sqrt{6}}{2} l n (\frac{\sqrt{6} + 1}{\sqrt{6} - 1}) .

Commented by maxmathsup by imad last updated on 09/May/19

i t s o n l y a t r y c h a n g . t a n (\frac{x}{2}) = t g i v e

I = \int_{0}^{1} \frac{x}{s i n x} d x = \int_{0}^{t a n (\frac{1}{2})} \frac{2 a r c t a n (t)}{\frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 2 \int_{0}^{t a n (\frac{1}{2})} \frac{a r c t a n (t)}{t} d t

l e t i n t r o d u c e t h e p a r a m e t r i c f u n c t i o n f (α) = \int_{0}^{t a n (\frac{1}{2})} \frac{a r c t a n (t α)}{t} d t

w e h a v e f^{^{'}} (α) = \int_{0}^{t a n (\frac{1}{2})} \frac{t}{t (1 + α^{2} t^{2})} d t = \int_{0}^{t a n (\frac{1}{2})} \frac{d t}{1 + α^{2} t^{2}}

=_{α t = u} \int_{0}^{α t a n (\frac{1}{2})} \frac{d u}{α (1 + u^{2})} = \frac{1}{α} {[a r c t a n (u)]}_{0}^{α t a n (\frac{1}{2})} = \frac{1}{α} a r c t a n (α t a n (\frac{1}{2})) \Rightarrow

f (α) = \int \frac{1}{α} a r c t a n (α t a n (\frac{1}{2})) d α + c \dots b e c o n t i n u e d \dots