Question

Show that the function satisfies the DE

Answer 1

\[\begin{align*}
I(a)&=\int\limits_0^\infty e^{-u^2}\cos(au)\mathrm du\\
\frac{\mathrm dI(a)}{\mathrm da}&=\int\limits_0^\infty \frac{\partial}{\partial a}e^{-u^2}\cos(au)\mathrm du\\
&=-a\int\limits_0^\infty e^{-u^2}\sin(au)\mathrm du\\
\\
\end{align*}\]

Answer 2

now what should i do ?

Answer 3

firstly $$\frac{dI}{da}=\int\limits_{0}^{\infty}-ue^{-u^2}\sin(au)du$$

when we consider the left side of the differential equation,

$$\begin{align*}2\frac{dI}{da}+aI &= \int\limits_{0}^{\infty}(-2u)e^{-u^2}\sin(au)du+\int \limits_{0}^{\infty}e^{-u^2}a\cos(au)du\\
&=\int\limits_0^\infty\frac{de^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\sin(au)}{du}du\\
&=\int \limits_0^\infty \frac{d\,e^{-u^2}\sin(au)}{du}\,du\\
&=\int\limits_0^\infty d\,e^{-u^2}\sin(au)\\
&=\left[e^{-u^2}\sin(au)\right]_0^{\infty}=0-0=0
\end{align*}$$

Answer 4

some kinda reverse integration by parts?

Answer 5

im not quite sure how you got your third line

Answer 6

can you explain how you got from
\[\begin{align*} \int\limits_{0}^{\infty}(-2u)e^{-u^2}\sin(au)du+\int \limits_{0}^{\infty}e^{-u^2}a\cos(au)du\\
\text {to}\\
=\int\limits_0^\infty\frac{de^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\sin(au)}{du}du\end{align*}\]
@BAdhi

Answer 7

we know that $$\frac{d\,e^{-u^2}}{du}=(-2u)\,e^{-u^2}$$ and
$$\frac{d\,\sin(au)}{du}=a\cos(au)$$
so,
$$\begin{align*}\int\limits_0^\infty (-2u)\,e^{-u^2}\sin(au)\,du+\int\limits_0^\infty e^{-u^2}a\cos(au)\,du&=\int\limits_0^\infty \frac{d\,e^{-u^2}}{du}\sin(au)\,du+\int\limits_0^\infty e^{-u^2}\frac{d\,\sin(au)}{du}\,du\\
&=\int\limits_0^\infty \underbrace{\frac{d\,e^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\,\sin(au)}{du}}_{(1)}\,du\end{align*}$$

(1) can be reduced from the differentiation of a multiplication, property
(i.e.) $$\frac{d\,uv}{dx}=u\frac{du}{dx}+v\frac{dv}{dx}$$

$$\frac{d\,\left(e^{-u^2}\sin(au)\right)}{du}=\frac{d\,e^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\,\sin(au)}{du}$$
(1) can be replaced with this,

$$\begin{align*}\int\limits_0^\infty (-2u)\,e^{-u^2}\sin(au)\,du+\int\limits_0^\infty e^{-u^2}a\cos(au)\,du&=\int\limits_0^\infty \frac{d\,e^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\,\sin(au)}{du}\,du \\
&=\int\limits_0^\infty\frac{d\,\left(e^{-u^2}\sin(au)\right)}{du}du\end{align*}$$ Is it clear enough for you?

Answer 8

yes

Answer 9

\[\begin{align*}
2\frac{\mathrm dI(a)}{\mathrm da}+aI(a)
&=\int\limits_0^\infty -2ue^{-u^2}\sin(au)\mathrm du+\int\limits_0^\infty ae^{-u^2}\cos(au)\mathrm du\\
&=\int\limits_0^\infty\frac{\mathrm d }{\mathrm d u}\left(e^{-u^2}\sin(au)\right)\mathrm du\qquad\text{(using the product rule)}\\
&=\left.e^{-u^2}\sin(au)\right|_0^\infty\\
&=0
\end{align*}\]

Answer 10

\[\begin{align*}
I(a)&=\int\limits_0^\infty e^{-u^2}\cos(au)\,\mathrm du\\
I(0)&=\int\limits_0^\infty e^{-u^2}\,\mathrm du\\
\frac{\sqrt \pi}2&=\int\limits_0^\infty e^{-u^2}\,\mathrm du\\
\text{let }u=v^{1/2}\\
\mathrm du=\frac{v^{-1/2}}2\,\mathrm dv\\
\frac{\sqrt \pi}2&=\int\limits_0^\infty e^{-v}\frac{v^{-1/2}\,\mathrm dv}2\\
{\sqrt \pi}&=\int\limits_0^\infty {v^{-1/2}e^{-v}\,\mathrm dv}\\
{\sqrt \pi}&=\Gamma(\tfrac12)\\
\end{align*}\]

Answer 11

im not sure what they mean by an expression for \(I(a)\)

Answer 12

http://mathhelpforum.com/calculus/197830-how-show-function-satisfies-differential-equation.html

Answer 13

@Sepeario yes i have tried differentiating \(I(a)\)

Answer 14

@UnkleRhaukus integrate the expression using integration by parts technique.then subst for I(0).

Answer 15

\[\begin{align*}
I(a)&=\int\limits_0^\infty e^{-u^2}\cos(au)\,\mathrm du\\
\text{let }u=v^{1/2}\\
\mathrm du=\frac{v^{-1/2}}2\,\mathrm dv\\
I(a)&=\int\limits_0^\infty e^{-v}\cos(a\sqrt v)\frac{v^{-1/2}\,\mathrm dv}2\\
\end{align*}\]

how do i integrate by parts, there are three terms?

do i break up the cos into e^... bits?

Answer 16

\[2\frac{ dI }{ da }+aI=0\\2 \frac{ dI }{ I }+a\space da=0\\2lnI+\frac{ a^2 }{ 2 }=C\]when \(\displaystyle a=0,\;I=\frac{\sqrt{\pi}}{2}\)\[2\ln \frac{ \sqrt{\pi} }{ 2 }+\frac{ 0^2 }{ 2 }=C\]\[2\ln I+\frac{a^2}{2}=2\ln\frac{\sqrt{\pi}}{2}\]\[\vdots\\\ln \left[\frac{ 4 }{ \pi }I^2(a)\right]=-\frac{ a^2 }{ 2 }\\I^2(a)=\frac{\pi}{4}\exp(-a^2/2)\]

Answer 17

Thankyou @sirm3d & @BAdhi \[\begin{align*}
2\frac{\mathrm dI(a)}{\mathrm da}+aI(a)&=0\\
2\frac{\mathrm dI(a)}{I(a)}+a\,\mathrm da&=0\\
2\int\frac{\mathrm dI(a)}{I(a)}+\int a\,\mathrm da&=0\\
2\ln |I(a)|+\frac{a^2}2&=c
\end{align*}\]

\[\begin{align*}
2\ln |I(0)|&=c\\
2\ln\frac{\sqrt\pi}2&=c\\
\ln\frac{\pi}4&=c\end{align*}\]

\[\begin{align*}
2\ln |I(a)|+\frac{a^2}2&=\ln\frac{\pi}4\\
\ln |I(a)|&=\ln{\frac{\sqrt\pi}2}-\frac{a^2}4\\
I(a)&={\frac{\sqrt\pi}2}e^{-\frac{a^2}4}
\end{align*}
\]