Question

\[ \textbf{(Separable) Differential Equations} \\
\ \text{Evaluate } \int_{0}^{\infty} e^{-t^2 - (9/t^2)} \; dt \]

Answer 1

\( \normalsize{ Hint \text{: Let} \\
\quad I(x) = \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \text{.} \\
\qquad \color{red}{ \text{Calculate } I ' (x) \text{ and find a differential equation for } I(x) } \\
\qquad \text{Use the standard integral } \int_{0}^{\infty} e^{-t^2} \; dt = \frac{\sqrt{\pi}}{2} \text{ to determine } I(0) \\
\qquad \text{Use this initial condition to solve for } I(x) \\
\qquad \text{Evaluate } I(3) \text{.}} \)

I've been looking at this problem for a while, but I cannot figure out how to calculate that derivative. :P

/first Q

Answer 2

But, I think I can handle the rest of the problem if I know how to calculate the I'(x)...

Answer 3

let\[ I(x)=\int_{0}^{\infty} f(x,t)\ \text{d}t\]\[I'(x)=\int_{0}^{\infty} \frac{\partial f}{\partial x} \text{d}t\]but there is some condition for this derivative

Answer 4

Let the Integral\[F(x)=\int_{c}^{\infty} f(x,t)\ \text{d}t\]be convergent when \(x \in [a,b]\) . let the partial derivative \(\frac{\partial f}{\partial x}\) be continuous in the 2 variables \(t,x\) when \(t>c\) and \(x \in [a,b]\) . and let the integral\[\int_{c}^{\infty} \frac{\partial f}{\partial x} \text{d}t\]converge uniformly on \([a,b]\). then \(F(x)\) has a derivative given by\[F'(x)=\int_{c}^{\infty} \frac{\partial f}{\partial x} \text{d}t\]

Answer 5

it was from my notes of the book :
advanced calculus - taylor angus & wiley fayez

Answer 6

\[ - \left (t^2 + {9 \over t^2}\right ) = - \left( t + {3 \over t}\right)^2 + 6\]
Let,
\[ \left( t + {3 \over t}\right) = u \\
du = \left( 1 - {3 \over t^2} \right) dt\]

Answer 7

\[ t^2 - ut + 3 = 0 \\
t = {u \pm \sqrt{u^2 - 12 }\over 2}\]
this is ugly

Answer 8

let's try some brute method\[ - \left (t^2 + {9 \over t^2}\right ) = - \left( t - {3 \over t}\right)^2 - 6\]

Answer 9

the limits of integration seems to change so badly
t -> inf, u->inf
t-> 0, u->-inf

Answer 10

the problems remains the same ... i thought i would get rid of the denominator t^2

Answer 11

Hmm...
Well, I tried using the partial derivative of the inside for x:
\[ \int_{0}^{\infty} \neg \frac{2x}{t^2} e^{-t^2 - (x/t)^2} dt \]
Which looks interesting, having the extra 1/t^2 involved now. I feel like maybe experiment has something there too, but I have to sleep for now. I'll go over this problem more tomorrow. Thanks for the help! :D

Answer 12

sure...Access this is beautiful !

Answer 13

Okay, I took some time earlier today and I think I finally got it. :D

Answer 14

\[ I(x) = \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \\ \\
\begin{align}
I'(x) &= \frac{d}{dx} \left( \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \right) \\
&= \int_{0}^{\infty} \frac{\partial}{\partial x} \left( e^{-t^2 - (x/t)^2} \right) \; dt \quad (t > 0) \\
&= \int_{0}^{\infty} \left( \neg \frac{2x}{t^2} \right) e^{-t^2 - (x/t)^2} \; dt \\
&= \int_{0}^{\infty} \left( \neg \frac{2x}{t^2} \right) e^{-(t - x/t)^2 - 2x} \; dt \\
&= \int_{0}^{\infty} \left( \neg \frac{2x}{t^2} \right) e^{-2x} e^{-(t - x/t)^2} \; dt \\
&= \neg 2e^{-2x} \int_{0}^{\infty} \frac{x}{t^2} e^{-(t - x/t)^2} \; dt \\
& \quad u = t - \frac{x}{t} \\
& \quad du = 1 + \frac{x}{t^2} \; dt \quad \text{Add/subtract to get the +1 coefficient} \\
& \qquad \implies \text{Upper Bound: } \lim_{t \to \infty} u = \infty \text{,} \\
& \qquad \implies \text{Lower Bound: } \lim_{t \to 0^{+}} u = - \infty \text{. (From:} t>0) \\
&= \neg 2e^{-2x} \int_{0}^{\infty} \left( \frac{x}{t^2} e^{-(t - x/t)^2} + e^{-(t - x/t)^2} - e^{-(t - x/t)^2} \right) \; dt \\
&= \neg 2e^{-2x} \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} - e^{-(t - x/t)^2} \; dt \\
&= \neg 2e^{-2x} \left( \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} \; dt - \int_{0}^{\infty} e^{-(t - x/t)^2} \; dt \right) \\
&= \neg 2e^{-2x} \int_{0}^{\infty} \left(1 + \frac{x}{t^2} \right) e^{-(t - x/t)^2} \; dt + 2e^{-2x} \int_{0}^{\infty} e^{-(t - x/t)^2} \; dt \\
&= \neg 2e^{-2x} \int_{-\infty}^{\infty} e^{-u^2} \; du + 2 \int_{0}^{\infty} e^{-t^2 - (x/t)^2} \; dt \\
I'(x) &= \neg 2e^{-2x} \sqrt{\pi} + 2 I(x)
\end{align}
\]

It 'seems' nice.. :P

Answer 15

Although, the resulting equation does not seem to be a separable DE... it looks more like a linear DE.

Answer 16

Yep .. linear in x

Answer 17

one way to do it
\[ Let, - \left (t^2 + {9 \over t^2}\right ) = - \left( t - {3 \over t}\right)^2 - 6 \\
\]
you have