let-f-x-0-arctan-xt-2-cos-t-2-dt-1-find-a-explicite-form-of-f-x-2-find-a-explicite-form-of-f-x-3-find-the-value-of-0-cos-t-2-arctan-t-2-dt-and-0-cos-t-2-arc

Question Number 42374 by maxmathsup by imad last updated on 24/Aug/18

l e t f (x) = \int_{0}^{\infty} a r c t a n ({x t}^{2}) c o s (t^{2}) d t

1) f i n d a e x p l i c i t e f o r m o f f^{^{'}} (x)

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

3) f i n d t h e v a l u e o f \int_{0}^{\infty} c o s (t^{2}) a r c t a n (t^{2}) d t a n d \int_{0}^{\infty} c o s (t^{2}) a r c t a n (2 t^{2}) d t

Commented by maxmathsup by imad last updated on 26/Aug/18

1) w e h a v e f (x) = \int_{0}^{\infty} a r c t a n ({x t}^{2}) c o s (t^{2}) d t \Rightarrow

f^{^{'}} (x) = \int_{0}^{\infty} \frac{t^{2}}{1 + x^{2} t^{4}} c o s (t^{2}) d t \Rightarrow 2 f^{^{'}} (x) = \int_{- \infty}^{+ \infty} \frac{t^{2} c o s (t^{2})}{1 + x^{2} t^{4}} d t

= R e (\int_{- \infty}^{+ \infty} \frac{t^{2} e^{{i t}^{2}}}{1 + x^{2} t^{4}}) d t l e t c o n s i d e r t h e c o m p l e x f u n c t i o n

φ (z) = \frac{z^{2} e^{{i z}^{2}}}{x^{2} z^{4} + 1} w e h a v e φ (z) = \frac{z^{2} e^{{i z}^{2}}}{({x z}^{2} - i) ({x z}^{2} + i)}

= \frac{z^{2} e^{{i z}^{2}}}{x^{2} (z^{2} - \frac{i}{x}) (z^{2} + \frac{i}{x})} i f x > 0 w e g e t

φ (z) = \frac{z^{2} e^{{i z}^{2}}}{x^{2} (z - \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z + \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) (z - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) (z + \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}})}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) + R e s (φ, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}})}

R e s (φ, \frac{1}{\sqrt{x}} e^{\frac{i π}{4}}) = \frac{\frac{1}{x} {i e}^{i \frac{1}{x} i}}{x^{2} (\frac{2}{\sqrt{x}} e^{\frac{i π}{4}}) (2 \frac{i}{x})} = \frac{\sqrt{x} e^{- \frac{1}{x}}}{4 x^{2} e^{\frac{i π}{4}}} = \frac{\sqrt{x}}{4 x^{2}} e^{- \frac{1}{x} - \frac{i π}{4}} = \frac{1}{4} \frac{e^{- \frac{1}{x}}}{x \sqrt{x}} e^{- \frac{i π}{4}}

R e s (φ, - \frac{1}{\sqrt{x}} e^{- \frac{i π}{4}}) = \frac{(\frac{- i}{x}) e^{i (- \frac{i}{x})}}{x^{2} (\frac{- 2 i}{x}) (\frac{- 2}{\sqrt{x}} e^{- \frac{i π}{4}})} = - \frac{1}{4} \frac{e^{\frac{1}{x}}}{x \sqrt{x}} e^{\frac{i π}{4}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{4 x \sqrt{x}} {e^{- \frac{1}{x} - \frac{i π}{4}} - e^{\frac{1}{x} + \frac{i π}{4}}} = - \frac{i π}{2 x \sqrt{x}} {e^{\frac{1}{x} + \frac{i π}{4}} - e^{- \frac{1}{x} - \frac{i π}{4}}}

= - \frac{i π}{2 x \sqrt{x}} {2 i I m (e^{\frac{1}{x} + \frac{i π}{4}})} = \frac{π}{x \sqrt{x}} \frac{1}{\sqrt{2}} e^{\frac{1}{x}} \Rightarrow 2 f^{^{'}} (x) = \frac{π}{x \sqrt{2 x}} e^{\frac{1}{x}} \Rightarrow

f^{^{'}} (x) = \frac{π}{2 x \sqrt{2 x}} e^{\frac{1}{x}} \Rightarrow f (x) = \int^{x} \frac{π}{2 t \sqrt{2 t}} e^{\frac{1}{t}} d t + c . \dots . b e c o n t i n u e d \dots