let f(a) = ∫_(−∞) ^(+∞) cos(ax^2 )dx with a>0 1) calculate f(a) interms of a ) calculate ∫_(−∞) ^(+∞) cos(2x^2 )dx 3) find the value of ∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 42260 by math khazana by abdo last updated on 21/Aug/18

$l e t f (a) = \int_{- \infty}^{+ \infty} c o s ({a x}^{2}) d x w i t h a > 0$ $1) c a l c u l a t e f (a) i n t e r m s o f a$ $) c a l c u l a t e \int_{- \infty}^{+ \infty} c o s (2 x^{2}) d x$ $3) f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} c o s (x^{2} + x + 1) d x .$
Commented bymaxmathsup by imad last updated on 21/Aug/18
$3) ∫_(−∞) ^(+∞) cos(x^2 +x+1)dx =∫_(−∞) ^(+∞) cos((x+(1/2))^2 +(3/4))dx =_(x+(1/2)=((√3)/2)t) ∫_(−∞) ^(+∞) cos((3/4)(t^2 +1))((√3)/2) dt =((√3)/2) ∫_(−∞) ^(+∞) cos((3/4)t^2 +(3/4))dt =((√3)/2) ∫_(−∞) ^(+∞) {cos((3/4)t^2 )cos((3/4))−sin((3/4)t^2 )sin((3/4))}dt =((√3)/2) cos((3/4)) ∫_(−∞) ^(+∞) cos((3/4)t^2 )dt −((√3)/2) sin((3/4))∫_(−∞) ^(+∞) sin((3/4)t^2 )dt =((√3)/2) cos((3/4))((√(2π))/(2(√(3/4)))) −((√3)/2) sin((3/4))((√(2π))/(2(√(3/4)))) =((√(2π))/2) cos((3/4)) −((√(2π))/2) sin((3/4)) ⇒ ∫_(−∞) ^(+∞) cos(x^2 +x +1)dx =(√(π/2)){cos((3/4))−sin((3/4))} .$
$3) \int_{- \infty}^{+ \infty} c o s (x^{2} + x + 1) d x = \int_{- \infty}^{+ \infty} c o s ({(x + \frac{1}{2})}^{2} + \frac{3}{4}) d x$ $=_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t} \int_{- \infty}^{+ \infty} c o s (\frac{3}{4} (t^{2} + 1)) \frac{\sqrt{3}}{2} d t$ $= \frac{\sqrt{3}}{2} \int_{- \infty}^{+ \infty} c o s (\frac{3}{4} t^{2} + \frac{3}{4}) d t$ $= \frac{\sqrt{3}}{2} \int_{- \infty}^{+ \infty} {c o s (\frac{3}{4} t^{2}) c o s (\frac{3}{4}) - s i n (\frac{3}{4} t^{2}) s i n (\frac{3}{4})} d t$ $= \frac{\sqrt{3}}{2} c o s (\frac{3}{4}) \int_{- \infty}^{+ \infty} c o s (\frac{3}{4} t^{2}) d t - \frac{\sqrt{3}}{2} s i n (\frac{3}{4}) \int_{- \infty}^{+ \infty} s i n (\frac{3}{4} t^{2}) d t$ $= \frac{\sqrt{3}}{2} c o s (\frac{3}{4}) \frac{\sqrt{2 π}}{2 \sqrt{\frac{3}{4}}} - \frac{\sqrt{3}}{2} s i n (\frac{3}{4}) \frac{\sqrt{2 π}}{2 \sqrt{\frac{3}{4}}}$ $= \frac{\sqrt{2 π}}{2} c o s (\frac{3}{4}) - \frac{\sqrt{2 π}}{2} s i n (\frac{3}{4}) \Rightarrow$ $\int_{- \infty}^{+ \infty} c o s (x^{2} + x + 1) d x = \sqrt{\frac{π}{2}} {c o s (\frac{3}{4}) - s i n (\frac{3}{4})} .$
Commented bymaxmathsup by imad last updated on 21/Aug/18
$we have ∫_(−∞) ^(+∞) cos(ax^2 )dx−i ∫_(−∞) ^(+∞) sin(ax^2 )dx = ∫_(−∞) ^(+∞) e^(−iax^2 ) dx =∫_(−∞) ^(+∞) e^(−((√(ia))x)^2 ) dx =_((√(ia))x =t) ∫_(−∞) ^(+∞) e^(−t^2 ) (dt/(√(ia))) = (1/((√a)e^((iπ)/4) )) (√π)=((√π)/(√a)) e^(−((iπ)/4)) =((√π)/(√a)) {cos((π/4))−isin((π/4))} ⇒ f(a) =((√π)/(√a)) ((√2)/2) =((√(2π))/(2(√a))) . also we have ∫_(−∞) ^(+∞) sin(ax^2 )dx =((√(2π))/(2(√a))) 2) ∫_(−∞) ^(+∞) cos(2x^2 )dx =f(2) = ((√(2π))/(2(√2))) =((√π)/2) .$
$w e h a v e \int_{- \infty}^{+ \infty} c o s ({a x}^{2}) d x - i \int_{- \infty}^{+ \infty} s i n ({a x}^{2}) d x$ $= \int_{- \infty}^{+ \infty} e^{- {i a x}^{2}} d x = \int_{- \infty}^{+ \infty} e^{- {(\sqrt{i a} x)}^{2}} d x =_{\sqrt{i a} x = t} \int_{- \infty}^{+ \infty} e^{- t^{2}} \frac{d t}{\sqrt{i a}}$ $= \frac{1}{\sqrt{a} e^{\frac{i π}{4}}} \sqrt{π} = \frac{\sqrt{π}}{\sqrt{a}} e^{- \frac{i π}{4}} = \frac{\sqrt{π}}{\sqrt{a}} {c o s (\frac{π}{4}) - i s i n (\frac{π}{4})} \Rightarrow$ $f (a) = \frac{\sqrt{π}}{\sqrt{a}} \frac{\sqrt{2}}{2} = \frac{\sqrt{2 π}}{2 \sqrt{a}} . a l s o w e h a v e \int_{- \infty}^{+ \infty} s i n ({a x}^{2}) d x = \frac{\sqrt{2 π}}{2 \sqrt{a}}$ $2) \int_{- \infty}^{+ \infty} c o s (2 x^{2}) d x = f (2) = \frac{\sqrt{2 π}}{2 \sqrt{2}} = \frac{\sqrt{π}}{2} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum