Integrate Gaussian Functions

←Older revision		Revision as of 18:01, 2 January 2017
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	#**<math>f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^{2}}{2\sigma^{2}}}</math>		#**<math>f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^{2}}{2\sigma^{2}}}</math>
	===Generalizations===		===Generalizations===
−	#Consider the integral below. The Gaussian integral <math>\int_{0}^{\infty}e^{-\alpha x^{2}}\mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}</math> is a result that can be used to find numerous related integrals. The ones below are called ''moments'' of the Gaussian. Below, <math>n</math> is a natural number.	+	#Consider the integral below. The Gaussian integral <math>\int_{0}^{\infty}e^{-\alpha x^{2}}\mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}</math> is a result that can be used to find numerous related integrals. The ones below are called ''moments'' of the Gaussian. Below, <math>n</math> is a positive number.
	#*<math>\int_{0}^{\infty}x^{n}e^{-x^{2}}\mathrm{d}x</math>		#*<math>\int_{0}^{\infty}x^{n}e^{-x^{2}}\mathrm{d}x</math>
−	#If <math>n</math> is odd, use the u-sub <math>u = x^{2}</math>. Then we can use the [[Integrate Using the Gamma Function|Gamma function]] to easily evaluate. Below, we choose <math>n = 9</math> as one example. Be sure to verify the general case.
−	#*<math>\begin{align}\int_{0}^{\infty}x^{9}e^{-x^{2}}\mathrm{d}x &= \frac{1}{2}\int_{0}^{\infty}u^{4}e^{-u}\mathrm{d}u \\
−	&= \frac{4!}{2} = 12\end{align}</math>
−	#*<math>\int_{0}^{\infty}x^{n}e^{-x^{2}}\mathrm{d}x = \frac{1}{2}\left(\frac{n-1}{2}\right)!,\ n = 1,3,5,\cdots</math>
	#If <math>n</math> is even, consider the related integral (written below) and [[Integrate by Differentiating Under the Integral|differentiate under the integral]]. The result from differentiating under the integral is that even powers of <math>x</math> get brought down. Notice that as the integral gets negated, the result on the right also gets negated because of the negative power in <math>\alpha,</math> so the answers remain positive. Since differentiation is much easier than integration, we could do this all day, making sure to set <math>\alpha = 1</math> at a convenient time. We list some of these integrals below. Make sure to verify them for yourself.		#If <math>n</math> is even, consider the related integral (written below) and [[Integrate by Differentiating Under the Integral|differentiate under the integral]]. The result from differentiating under the integral is that even powers of <math>x</math> get brought down. Notice that as the integral gets negated, the result on the right also gets negated because of the negative power in <math>\alpha,</math> so the answers remain positive. Since differentiation is much easier than integration, we could do this all day, making sure to set <math>\alpha = 1</math> at a convenient time. We list some of these integrals below. Make sure to verify them for yourself.
	#*<math>\int_{0}^{\infty}e^{-\alpha x^{2}}\mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}</math>		#*<math>\int_{0}^{\infty}e^{-\alpha x^{2}}\mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}</math>
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	#*<math>\int_{0}^{\infty}x^{4}e^{-x^{2}}\mathrm{d}x = \frac{3\sqrt{\pi}}{8}</math>		#*<math>\int_{0}^{\infty}x^{4}e^{-x^{2}}\mathrm{d}x = \frac{3\sqrt{\pi}}{8}</math>
	#*<math>\int_{0}^{\infty}x^{6}e^{-x^{2}}\mathrm{d}x = \frac{15\sqrt{\pi}}{16}</math>		#*<math>\int_{0}^{\infty}x^{6}e^{-x^{2}}\mathrm{d}x = \frac{15\sqrt{\pi}}{16}</math>
		+	#If <math>n</math> is not even, use the u-sub <math>u = x^{2}</math>. Then we can use the [[Integrate Using the Gamma Function|Gamma function]] to easily evaluate. Below, we choose <math>n = 9</math> and <math>n = 1/3</math> as examples.
		+	#*<math>\int_{0}^{\infty}x^{9}e^{-x^{2}}\mathrm{d}x = \frac{1}{2}\int_{0}^{\infty}u^{4}e^{-u}\mathrm{d}u = \frac{4!}{2} = 12</math>
		+	#*<math>\int_{0}^{\infty}x^{1/3}e^{-x^{2}}\mathrm{d}x = \frac{1}{2}\int_{0}^{\infty}u^{-1/3}e^{-u}\mathrm{d}u = \frac{\Gamma(2/3)}{2}</math>
	#Set <math>\alpha = i</math> to obtain three integrals. The result is general enough such that <math>\alpha</math> can even take on complex values. Recall Euler's formula relating the complex exponential function to the trigonometric functions. If we take the real and imaginary parts of our result, we obtain two integrals for free. Neither of the two real integrals have antiderivatives that can be written in closed form.		#Set <math>\alpha = i</math> to obtain three integrals. The result is general enough such that <math>\alpha</math> can even take on complex values. Recall Euler's formula relating the complex exponential function to the trigonometric functions. If we take the real and imaginary parts of our result, we obtain two integrals for free. Neither of the two real integrals have antiderivatives that can be written in closed form.
	#*<math>e^{-ix^{2}} = \cos x^{2} - i\sin x^{2}</math>		#*<math>e^{-ix^{2}} = \cos x^{2} - i\sin x^{2}</math>
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	#*<math>\int_{0}^{\infty}\cos x^{2}\mathrm{d}x = \int_{0}^{\infty}\sin x^{2}\mathrm{d}x = \frac{\sqrt{\pi}}{2\sqrt{2}}</math>		#*<math>\int_{0}^{\infty}\cos x^{2}\mathrm{d}x = \int_{0}^{\infty}\sin x^{2}\mathrm{d}x = \frac{\sqrt{\pi}}{2\sqrt{2}}</math>
	#*These two integrals are special cases of the '''Fresnel integrals,''' where they are important in the study of optics.		#*These two integrals are special cases of the '''Fresnel integrals,''' where they are important in the study of optics.
−	#Calculate the [[Calculate the Fourier Transform of a Function|Fourier transform]] of the Gaussian function. Calculating the Fourier transform is computationally very simple, but it requires a slight modification.	+	#Calculate the [[Calculate the Fourier Transform of a Function|Fourier transform]] of the Gaussian function by completing the square. Calculating the Fourier transform is computationally very simple, but it requires a slight modification. We opt to complete the square because we recognize the property that the integral is ''independent'' of the shift (see the tips). Since we have to add 0 in order to not change the integrand, we have to compensate by adding a <math>-\frac{\omega^{2}}{4}</math> term.
−	#*<math>\mathcal{F}\{e^{-t^{2}}\} = \int_{-\infty}^{\infty}e^{-t^{2}}e^{-i\omega t}\mathrm{d}t</math>	+
−	#Complete the square in the exponent. We opt to complete the square because we recognize the property that the integral is ''independent'' of the shift (see the tips). Since we have to add 0 in order to not change the integrand, we have to compensate by adding a <math>-\frac{\omega^{2}}{4}</math> term.	+
	#*<math>\begin{align}\mathcal{F}\{e^{-t^{2}}\} &= \int_{-\infty}^{\infty}e^{-t^{2}}e^{-i\omega t}\mathrm{d}t \\		#*<math>\begin{align}\mathcal{F}\{e^{-t^{2}}\} &= \int_{-\infty}^{\infty}e^{-t^{2}}e^{-i\omega t}\mathrm{d}t \\
	&= \int_{-\infty}^{\infty}e^{-(t^{2}+i\omega t - \omega^{2}/4 + \omega^{2}/4)}\mathrm{d}t \\		&= \int_{-\infty}^{\infty}e^{-(t^{2}+i\omega t - \omega^{2}/4 + \omega^{2}/4)}\mathrm{d}t \\

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