My Math Forum ODE & Improper Integral

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 December 4th, 2011, 10:54 AM #1 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 ODE & Improper Integral Can someone please help me with this integral $\int_0^\infty e^{-x^2}sinx\,\mathrm{d}x$? I put F(K)=$\int_0^\infty e^{-x^2}sinx\,\mathrm{d}x$ but I have problem with the F(0).
 December 4th, 2011, 10:55 AM #2 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 Re: ODE & Improper Integral Sorry, I put F(k)=$\int_0^\infty e^{-x^2}sinkx\,\mathrm{d}x$ but I stuck everytime in F(0).
 December 4th, 2011, 02:04 PM #3 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: ODE & Improper Integral $\int e^{-x^2} \sin (k x) \, dx=\int e^{-x^2} \frac{\left(e^{ikx}-e^{-ikx}\right)}{2i}dx$ $=\frac{1}{2i}\int e^{-x^2+\text{ikx}}-e^{-x^2-\text{ikx}}dx$ $=\frac{1}{2i}\int e^{-\left(x-\frac{\text{ik}}{2}\right)^2-\frac{k^2}{4}}-e^{-\left(x+\frac{\text{ik}}{2}\right)^2-\frac{k^2}{4}}dx$ $=\frac{\sqrt{\pi }}{4i}e^{\frac{-k^2}{4}}\left(\text{Erf}\left(x-\frac{\text{ik}}{2}\right)-\text{Erf}\left(x+\frac{\text{ik}}{2}\right)\right )$ $\int_0^\infty e^{-x^2} \sin (k x) \, dx=\lim_{x \to \infty }\frac{\sqrt{\pi }}{4i}e^{\frac{-k^2}{4}}\left(\text{Erf}\left(x-\frac{\text{ik}}{2}\right)-\text{Erf}\left(x+\frac{\text{ik}}{2}\right)\right ) - \frac{\sqrt{\pi }}{4i}e^{\frac{-k^2}{4}}\left(\text{Erf}\left(-\frac{\text{ik}}{2}\right)-\text{Erf}\left(\frac{\text{ik}}{2}\right)\right)$ $=\lim_{x \to \infty }\frac{\sqrt{\pi }}{4i}e^{\frac{-k^2}{4}}\left(\text{Erf}\left(x-\frac{\text{ik}}{2}\right)-\text{Erf}\left(x+\frac{\text{ik}}{2}\right)\right ) + \frac{\sqrt{\pi }}{2i}e^{\frac{-k^2}{4}}\left(\text{Erf}\left(\frac{\text{ik}}{2}\ right)\right)$ $=\lim_{x \to \infty }\frac{\sqrt{\pi }}{4i}e^{\frac{-k^2}{4}}\left(\text{Erf}\left(x-\frac{\text{ik}}{2}\right)-\text{Erf}\left(x+\frac{\text{ik}}{2}\right)\right ) + \frac{\sqrt{\pi }}{2}e^{\frac{-k^2}{4}}\left(\text{Erfi}\left(\frac{\text{k}}{2}\ right)\right)$ $=\lim_{x \to \infty }\frac{\sqrt{\pi }}{4i}e^{\frac{-k^2}{4}}\left(\text{Erf}\left(x-\frac{\text{ik}}{2}\right)-\text{Erf}\left(x+\frac{\text{ik}}{2}\right)\right ) + \frac{\sqrt{\pi }}{2}e^{\frac{-k^2}{4}}\left(\text{Erfi}\left(\frac{\text{k}}{2}\ right)\right)$ $=DawsonF(\frac{k}{2})$ $\lim_{x \to \infty }\frac{\sqrt{\pi }}{4i}e^{\frac{-k^2}{4}}\left(\text{Erf}\left(x-\frac{\text{ik}}{2}\right)-\text{Erf}\left(x+\frac{\text{ik}}{2}\right)\right )=0$ because integrand goes to 0 at infinity + analytic function
 December 5th, 2011, 05:19 AM #4 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 Re: ODE & Improper Integral I must solve the improper equation by using Ordinary Differential Equations, but I can't here is what I do: $F(k)$=$\int_0^\infty e^{-x^2}sinkx\,\mathrm{d}x$ $dF/dk$$=$$\int_0^\infty xe^{-x^2}coskx\,\mathrm{d}x$$=$$-1/2$$\int_0^\infty -2xe^{-x^2}coskx\,\mathrm{d}x$$=$$-1/2$$\int_0^\infty (e^{-x^2})'coskx\,\mathrm{d}x$=$[(-1/2)e^{-x^2}coskx]_0^\infty$$-k/2$$\int_0^\infty e^{-x^2}sinkx\,\mathrm{d}x$$=$$(-k/2)F(k)$ Thus: $dF/dk$$=$$(-k/2)F(k)$ I solve the differential equation and then: $F(k)=ce^{(-k^2)/4}$ to find $c$ I put $k=0$ and there I have problem. How I can solve the improper integral using ODE and If it can't be solved by using ODE how I can solve it using complex analysis?
 December 5th, 2011, 06:29 AM #5 Senior Member   Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: ODE & Improper Integral I think it should be $\frac{\mathrm{d} F}{\mathrm{d} k}=\frac{1}{2}-\frac{k}{2}F$ $F(0)=0$
December 5th, 2011, 06:45 AM   #6
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Re: ODE & Improper Integral

Quote:
 Originally Posted by wnvl I think it should be $\frac{\mathrm{d} F}{\mathrm{d} k}=\frac{1}{2}-\frac{k}{2}F$ $F(0)=0$
Checked this ODE and the solution is indeed again $=DawsonF(\frac{k}{2})$.

 December 5th, 2011, 07:04 AM #7 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 Re: ODE & Improper Integral why it should be $dF/dk=(1/2)-k/2F$ where I did wrong to the integral? But mostly why $F(0)$? if $k=0$ then the integral becomes $\int_0^\infty 0\,\mathrm{d}x$ which is not .
December 5th, 2011, 07:15 AM   #8
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Re: ODE & Improper Integral

Quote:
 Originally Posted by talisman why it should be $dF/dk=(1/2)-k/2F$ where I did wrong to the integral? But mostly why $F(0)$? if $k=0$ then the integral becomes $\int_0^\infty 0\,\mathrm{d}x$ which is not .
you made an error in the evaluation of $[(-1/2)e^{-x^2}coskx]_0^\infty$ for 0

concerning your second question, when the integrand is zero over the entire range of integration then the integral should be zero.

 December 5th, 2011, 07:44 AM #9 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 Re: ODE & Improper Integral Can you explain how I will find the $[(-1/2)e^{-x^2}coskx]_0^\infty$? and also why the integral of zero over the entire range of integration is zero? I asked 2 teachers in my university and told me that the integral of zero can not be defined.
December 5th, 2011, 08:04 AM   #10
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Re: ODE & Improper Integral

Quote:
 Originally Posted by talisman Can you explain how I will find the $[(-1/2)e^{-x^2}coskx]_0^\infty$? and also why the integral of zero over the entire range of integration is zero? I asked 2 teachers in my university and told me that the integral of zero can not be defined.

$[(-1/2)e^{-0^2}cosk0]=(-1/2)*1*1=-1/2$
$[(-1/2)e^{-\infty^2}cosk\infty]=(-1/2)*0*(something between -1 and 1)=0$

Quote:
 Originally Posted by talisman and also why the integral of zero over the entire range of integration is zero? I asked 2 teachers in my university and told me that the integral of zero can not be defined.
For me this is trivial, but I don't have the background to give a strict proof.

 Tags improper, integral, ode

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