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Question Number 48261 by Abdo msup. last updated on 21/Nov/18

l e t f (x) = \int_{\frac{1}{2}}^{1} \frac{d t}{2 + c h (x t)}

1) f i n d a e x p l i c i t f o r m o f f (x)

2) c a l c u l a t e g (x) = \int_{\frac{1}{2}}^{1} \frac{t s h (x t)}{{(2 + c h (x t))}^{2}} d t

3) f i n d t h e v a l u e o f \int_{\frac{1}{2}}^{1} \frac{d t}{2 + c h (3 t)} a n d \int_{\frac{1}{2}}^{1} \frac{t s h (2 t)}{{(2 + c h (2 t))}^{2}} d t

4) l e t u_{n} = \int_{\frac{1}{2}}^{1} \frac{d t}{2 + c h (n t)} s t u d y t h e c o n v e r g e n c e o f Σ u_{n}

a n d Σ \frac{u_{n}}{n} .

Commented by maxmathsup by imad last updated on 22/Nov/18

1) w e h a v e f o r x \neq 0 f (x) = \int_{\frac{1}{2}}^{1} \frac{d t}{2 + c h (x t)} =_{x t = u} \int_{\frac{x}{2}}^{x} \frac{d u}{x (2 + c h (u)}

= \frac{1}{x} \int_{\frac{x}{2}}^{x} \frac{d u}{2 + \frac{e^{u} + e^{- u}}{2}} = \frac{2}{x} \int_{\frac{x}{2}}^{x} \frac{d u}{4 + e^{u} + e^{- u}} =_{e^{u} = α} \frac{2}{x} \int_{e^{\frac{x}{2}}}^{e^{x}} \frac{d α}{α (4 + α + α^{- 1})}

= \frac{2}{x} \int_{e^{\frac{x}{2}}}^{e^{x}} \frac{d α}{α^{2} + 4 α + 1} = \frac{2}{x} \int_{e^{\frac{x}{2}}}^{e^{x}} \frac{d α}{{(α + 2)}^{2} - 3} = \frac{2}{x} \int_{e^{\frac{x}{2}}}^{e^{x}} \frac{d α}{(α - 1) (α + 5)}

= \frac{1}{3 x} \int_{e^{\frac{x}{2}}}^{e^{x}} {\frac{1}{α - 1} - \frac{1}{α + 5}} d α = \frac{1}{3 x} {[l n ∣ \frac{α - 1}{α + 5} ∣]}_{e^{\frac{x}{2}}}^{e^{x}}

= \frac{1}{3 x} {l n ∣ \frac{e^{x} - 1}{e^{x} + 5} ∣ - l n ∣ \frac{e^{\frac{x}{2}} - 1}{e^{\frac{x}{2}} + 5} ∣}

Commented by maxmathsup by imad last updated on 22/Nov/18

2) w e h a v e f^{^{'}} (x) = \int_{\frac{1}{2}}^{1} \frac{\partial}{\partial x} (\frac{d t}{2 + c h (x t)}) = \int_{\frac{1}{2}}^{1} \frac{- t s h (x t)}{{(2 + c h (x t))}^{2}} d t \Rightarrow

\int_{\frac{1}{2}}^{1} \frac{t s h (x t)}{(2 + c h {(x t)}^{2})} d t = - f^{^{'}} (x) b u t

f^{^{'}} (x) = - \frac{1}{3 x^{2}} {l n ∣ \frac{e^{x} - 1}{e^{x} + 5} ∣ - l n ∣ \frac{e^{\frac{x}{2}} - 1}{e^{\frac{x}{2}} + 5} ∣

+ \frac{1}{3 x} {\frac{e^{x}}{e^{x} - 1} - \frac{e^{x}}{e^{x} + 5} - \frac{1}{2} \frac{e^{\frac{x}{2}}}{e^{\frac{x}{2} - 1}} + \frac{1}{2} \frac{e^{\frac{x}{2}}}{e^{\frac{x}{2}} + 5}} \Rightarrow g (x) = - f^{^{'}} (x)

Commented by maxmathsup by imad last updated on 22/Nov/18

3) \int_{\frac{1}{2}}^{1} \frac{d t}{2 + c h (3 t)} = f (3) = \frac{1}{9} {l n ∣ \frac{e^{3} - 1}{e^{3} + 5} ∣ - l n ∣ \frac{e^{\frac{3}{2}} - 1}{e^{\frac{3}{2}} + 5} ∣

Commented by maxmathsup by imad last updated on 22/Nov/18

\int_{\frac{1}{2}}^{1} \frac{t s h (2 t)}{{(2 + c h (2 t))}^{2}} d t = g (2) = - f^{^{'}} (2) = \frac{1}{12} {l n ∣ \frac{e^{2} - 1}{e^{2} + 5} ∣ - l n ∣ \frac{e - 1}{e + 5} ∣}

+ \frac{1}{6} {\frac{e^{2}}{e^{2} - 1} - \frac{e^{2}}{e^{2} + 5} - \frac{1}{2} \frac{e}{e - 1} + \frac{1}{2} \frac{e}{e + 5}}