calculate-f-x-0-e-xt-2-4-t-2-dt-with-x-gt-0-

Question Number 72988 by mathmax by abdo last updated on 05/Nov/19

c a l c u l a t e f (x) = \int_{0}^{\infty} \frac{e^{- {x t}^{2}}}{4 + t^{2}} d t w i t h x > 0

Commented by mathmax by abdo last updated on 05/Nov/19

w e h a v e f (x) = \int_{0}^{\infty} \frac{e^{- x (t^{2} + 4 - 4)}}{t^{2} + 4} d t = e^{4 x} \int_{0}^{\infty} \frac{e^{- x (t^{2} + 4)}}{t^{2} + 4} d t

= e^{4 x} w (x) w i t h w (x) = \int_{0}^{\infty} \frac{e^{- x (t^{2} + 4)}}{t^{2} + 4} d t \Rightarrow

w^{^{'}} (x) = - \int_{0}^{+ \infty} e^{- x (t^{2} + 4)} d t = - e^{- 4 x} \int_{0}^{\infty} e^{- {(\sqrt{x} t)}^{2}} d t

=_{\sqrt{x} t = u} - e^{- 4 x} \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{\sqrt{x}} = - \frac{e^{- 4 x}}{\sqrt{x}} \times \frac{\sqrt{π}}{2} \Rightarrow w (x) = - \frac{\sqrt{π}}{2} \int_{0}^{x} \frac{e^{- 4 u}}{\sqrt{u}} d u + c

=_{\sqrt{u} = z} - \frac{\sqrt{π}}{2} \int_{0}^{\sqrt{x}} \frac{e^{- 4 z^{2}}}{z} (2 z) d z + c = - \sqrt{π} \int_{0}^{\sqrt{x}} e^{- 4 z^{2}} d z + c \Rightarrow

f (x) = c e^{4 x} - \sqrt{π} e^{4 x} \int_{0}^{\sqrt{x}} e^{- 4 z^{2}} d z

c = {l i m}_{x \to 0} f (x) = \int_{0}^{\infty} \frac{d t}{t^{2} + 4} =_{t = 2 u} \int_{0}^{\infty} \frac{2 d u}{4 (1 + u^{2})} = \frac{1}{2} \frac{π}{2} = \frac{π}{4} \Rightarrow

f (x) = \frac{π}{4} e^{4 x} - \sqrt{π} e^{4 x} \int_{0}^{\sqrt{x}} e^{- 4 z^{2}} d z

Answered by mind is power last updated on 05/Nov/19

f^{'} (x) = \int_{0}^{+ \infty} \frac{- t^{2} e^{- {xt}^{2}}}{4 + t^{2}} = \int - e^{- {xt}^{2}} + 4 \int_{0}^{+ \infty} \frac{e^{- {xt}^{2}}}{4 + t^{2}} dt

=\Rightarrow f^{'} (x) = 4 f (x) - \frac{1}{\sqrt{x}} \int_{0}^{+ \infty} e^{- {(t \sqrt{x})}^{2}} . \sqrt{x} dt

\Rightarrow f^{'} (x) = 4 f (x) - \frac{1}{\sqrt{x}} . \int_{0}^{+ \infty} e^{- u^{2}} du = 4 f (x) - \frac{1}{2 \sqrt{x}} . \sqrt{\frac{π}{2}}

f^{'} (x) - 4 f (x) + \frac{1}{2 \sqrt{x}} . \sqrt{\frac{π}{2}} = 0

f (x) = {ke}^{4 x}

\Rightarrow k^{'} (x) = - \frac{e^{- 4 x}}{2 \sqrt{x}} \sqrt{\frac{π}{2}}

=\Rightarrow k (x) = - \frac{\sqrt{π}}{2 \sqrt{2}} \int \frac{e^{- 4 x}}{2 \sqrt{x}} dx

\erf (x) = \int_{0}^{x} e^{- t^{2}} dt

\sqrt{x} = u \Rightarrow k (x) = - \frac{\sqrt{π}}{2 \sqrt{2}} \int e^{- {4 u}^{2}} du = - \frac{\sqrt{π}}{4 \sqrt{2}} \int e^{- w^{2}} dw = - \frac{\sqrt{π}}{4 \sqrt{2}} \erf (w)

= \frac{\sqrt{π}}{4 \sqrt{2}} \erf (2 u) = - \frac{\sqrt{π}}{4 \sqrt{2}} \erf (2 \sqrt{x}) + c

\Rightarrow f (x) = {ce}^{- 4 x} - \frac{\sqrt{π}}{4 \sqrt{2}} \erf (2 \sqrt{x}) e^{- 4 x}

f (0) = \int_{0}^{+ \infty} \frac{1}{4 + t^{2}} = [\frac{1}{2} . \arctan (\frac{t}{2})] = \frac{π}{4}

\Rightarrow c = \frac{π}{4}

f (x) = \frac{π}{4} e^{- 4 x} - \frac{\sqrt{π}}{4 \sqrt{2}} \erf (2 \sqrt{x}) e^{- 4 x}