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Question Number 33883 by math khazana by abdo last updated on 26/Apr/18

f i n d a s i m p l e f o r m o f f (x) = \int_{0}^{\frac{π}{2}} l n (1 + {x s i n}^{2} t) d t

w i t h ∣ x ∣< 1 .

Commented by math khazana by abdo last updated on 01/May/18

w e h a v e f^{^{'}} (x) = \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} t}{1 + {x s i n}^{2} t} d t a n d f o r x \neq 0

f^{^{'}} (x) = \frac{1}{x} \int_{0}^{\frac{π}{2}} \frac{1 + {x s i n}^{2} t - 1}{1 + {x s i n}^{2} t} d t

= \frac{π}{2 x} - \frac{1}{x} \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x {s i n}^{2} t} b u t

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

= \int_{0}^{\frac{π}{2}} \frac{2 d t}{2 + x - x c o s (2 t)} =_{2 t = u} \int_{0}^{π} \frac{d u}{2 + x - x c o s u}

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

\int_{0}^{\frac{π}{2}} \frac{d t}{1 + x {s i n}^{2} t} = \int_{0}^{+ \infty} \frac{1}{2 + x - x \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

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

= \int_{0}^{\infty} \frac{2 d t}{2 + x + (2 + x) t^{2} - x + {x t}^{2}}

= \int_{0}^{\infty} \frac{2 d t}{2 + 2 (1 + x) t^{2}} = \int_{0}^{\infty} \frac{d t}{1 + (1 + x) t^{2}}

=_{\sqrt{1 + x} t = u} \int_{0}^{\infty} \frac{1}{1 + u^{2}} \frac{d u}{\sqrt{1 + x}} = \frac{π}{2 \sqrt{1 + x}} \Rightarrow

f^{^{'}} (x) = \frac{π}{2 x} - \frac{1}{x} \frac{π}{2 \sqrt{1 + x}} = \frac{π}{2 x} (1 - \frac{1}{\sqrt{1 + x}}) \Rightarrow

f (x) = \frac{π}{2} l n ∣ x ∣ - \frac{π}{2} \int_{.}^{x} \frac{d t}{t \sqrt{1 + t}} + λ

\int \frac{d t}{t \sqrt{1 + t}} =_{\sqrt{1 + t} = u} \int \frac{2 u d u}{(u^{2} - 1) u}

= \int \frac{2 d u}{(u + 1) (u - 1)} = \int (\frac{1}{u - 1} - \frac{1}{u + 1}) d u

= l n ∣ \frac{u - 1}{u + 1} ∣ = l n (\frac{\sqrt{1 + x} - 1}{\sqrt{1 + x} + 1}) \Rightarrow

f (x) = \frac{π}{2} l n ∣ x ∣ - \frac{π}{2} l n (\frac{\sqrt{1 + x} - 1}{\sqrt{1 + x} + 1}) + λ

λ = {l i m}_{x \to 1} (f (x) - \frac{π}{2} l n ∣ x ∣ + \frac{π}{2} l n (\frac{\sqrt{1 + x} - 1}{\sqrt{1 + x} + 1}))

= f (1) + \frac{π}{2} l n (\frac{\sqrt{2} - 1}{\sqrt{2} + 1})

= \int_{0}^{\frac{π}{2}} l n (1 + {s i n}^{2} t) d t + \frac{π}{2} l n (\frac{\sqrt{2} - 1}{\sqrt{2} + 1}) .