calculate-f-a-0-e-x-2-a-x-2-dx-with-a-gt-0-

Question Number 79107 by mathmax by abdo last updated on 22/Jan/20

c a l c u l a t e f (a) = \int_{0}^{\infty} e^{- (x^{2} + \frac{a}{x^{2}})} d x w i t h a > 0

Commented by mathmax by abdo last updated on 23/Jan/20

f (a) = \int_{0}^{\infty} e^{- (x^{2} + \frac{a}{x^{2}})} d x c h a n g e m e n t x = \frac{1}{t} g i v e

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

\Rightarrow f^{^{'}} (a) = - \int_{0}^{\infty} e^{- ({a t}^{2} + \frac{1}{t^{2}})} d t =_{\sqrt{a} t = x} - \frac{1}{\sqrt{a}} \int_{0}^{\infty} e^{- {x^{2} + \frac{a}{x^{2}}}} d x

= - \frac{1}{\sqrt{a}} f (a) \Rightarrow \frac{f^{^{'}} (a)}{f (a)} = - \frac{1}{\sqrt{a}} \Rightarrow l n ∣ f (a) ∣= - 2 \sqrt{a} + c

f > 0 \Rightarrow f (a) = k e^{- 2 \sqrt{a}}

k = f (0) = \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2} \Rightarrow f (x) = \frac{\sqrt{π}}{2} e^{- 2 \sqrt{a}}

Answered by ~blr237~ last updated on 22/Jan/20

l e t s t a t e u = \frac{1}{x} t h e n d x = - \frac{d u}{u^{2}}

f (a) = \int_{0}^{\infty} e^{- (\frac{1}{u^{2}} + {a u}^{2})} \frac{d u}{u^{2}}

f \in C^{1} c a u s e f o r a l l a > 0 ∣ e^{- (x^{2} + \frac{a}{x^{2}})} ∣⩽ e^{- x^{2}} a n d \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2} s o c o n v e r g e s

N o w w e c a n w r i t e

\frac{d f}{d a} = - \int_{0}^{\infty} e^{- ({a u}^{2} + \frac{1}{u^{2}})} d u

= - \frac{1}{\sqrt{a}} \int_{0}^{\infty} e^{- [{(u \sqrt{a})}^{2} + \frac{a}{{(u \sqrt{a})}^{2}}]} d (u \sqrt{a})

= - \frac{1}{\sqrt{a}} \int_{0}^{\infty} e^{- (v^{2} + \frac{a}{v^{2}})} d v w i t h v = u \sqrt{a}

S o \frac{d f}{d a} = - \frac{1}{\sqrt{a}} f (a)

T h e n l n ∣ f (a) ∣= - 2 \sqrt{a} + c \Rightarrow f (a) = {k e}^{- 2 \sqrt{a}} w i t h k > 0

w e k n o w t h a t \lim_{a \to 0} f (a) = k = \frac{\sqrt{π}}{2}

F i n a l y f (a) = \frac{\sqrt{π}}{2} e^{- 2 \sqrt{a}}

Commented by msup trace by abdo last updated on 22/Jan/20

t h a n k s s i r

Answered by mind is power last updated on 22/Jan/20

o n p o s e x = a^{\frac{1}{4}} u

\Rightarrow f (a) = \int_{0}^{+ \infty} a^{\frac{1}{4}} e^{- \sqrt{a} (x^{2} + \frac{1}{x^{2}})} d x

x = \frac{1}{y} \Rightarrow

= a^{\frac{1}{4}} \int_{0}^{+ \infty} e^{- \sqrt{a} (y^{2} + \frac{1}{y^{2}})} \frac{d y}{y^{2}}

\Rightarrow 2 f (a) a^{- \frac{1}{4}} = \int_{0}^{+ \infty} (1 + \frac{1}{y^{2}}) e^{- \sqrt{a} {(y - \frac{1}{y})}^{2} - 2 \sqrt{a}}

\Rightarrow 2 f (a) a^{- \frac{1}{4}} e^{2 \sqrt{a}} = \int_{0}^{+ \infty} (1 + \frac{1}{y^{2}}) e^{- \sqrt{a} {(y - \frac{1}{y})}^{2}} d y

= \int_{0}^{1} (1 + \frac{1}{y^{2}}) e^{- \sqrt{a} {(y - \frac{1}{y})}^{2}} + \int_{1}^{+ \infty} (1 + \frac{1}{y^{2}}) e^{- \sqrt{a} {(y - \frac{1}{y})}^{2}} d y

u = y - \frac{1}{y}, i n f i r s t t = \frac{1}{y} i n 2 n d \Rightarrow

\Rightarrow \int_{0}^{+ \infty} e^{- \sqrt{a} u^{2}} d u + \int_{0}^{1} (1 + \frac{1}{t^{2}}) e^{- \sqrt{a} {(t - \frac{1}{t})}^{2}} d t

= 2 \int_{0}^{+ \infty} e^{- \sqrt{a} (u^{2})} d u, w = a^{\frac{1}{4}} u \Rightarrow

= 2 \int_{0}^{+ \infty} e^{- w^{2}} \frac{d w}{a^{\frac{1}{4}}} = \frac{\sqrt{π}}{a^{\frac{1}{4}}}

\Rightarrow 2 f (a) a^{- \frac{1}{4}} e^{2 \sqrt{a}} = \frac{\sqrt{π}}{a^{\frac{1}{4}}} \Rightarrow f (a) = \frac{\sqrt{π}}{2 e^{2 \sqrt{a}}}

Commented by msup trace by abdo last updated on 22/Jan/20

t h a n k s s i r .

Commented by mind is power last updated on 22/Jan/20

y^{'} r e w e l c o m