Solve Gaussian Integral: Even/Odd Cases

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SUMMARY

The discussion focuses on solving the Gaussian integral \( I_n = \int_{0}^{\infty} x^n e^{-\lambda x^2} dx \) for both even and odd cases of \( n \). The user successfully evaluated the integral for \( n = 0 \) and derived the result for the modified Bessel function \( I_0 = \frac{1}{2} \sqrt{\frac{\pi}{\lambda}} \). The conversation highlights the formulas for even and odd cases, specifically \( I_{2n} = \frac{(2n-1)!}{2^{2n+1} \lambda^n} \sqrt{\frac{\pi}{\lambda}} \) and \( I_{2n+1} = \frac{n!}{2 \lambda^{n+1}} \), providing a clear distinction between the two scenarios.

PREREQUISITES
  • Understanding of Gaussian integrals
  • Familiarity with the properties of even and odd functions
  • Knowledge of the modified Bessel function of the first kind
  • Basic calculus, including integration techniques
NEXT STEPS
  • Study the derivation of the Gaussian integral for various values of \( n \)
  • Learn about the modified Bessel function and its applications in integrals
  • Explore the relationship between derivatives of Bessel functions and Gaussian integrals
  • Investigate piecewise functions and their applications in mathematical analysis
USEFUL FOR

Students in mathematics, particularly those studying calculus and integral equations, as well as researchers and educators looking to deepen their understanding of Gaussian integrals and their properties.

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EDIT: meant to post this is the math forums, if you can remove this I'm going to switch it over

Homework Statement


Solve:

In = \int_{0}^{\infty} x^n e^{-\lambda x^2} dx

Homework Equations


The Attempt at a Solution


So my teacher gave a few hints regarding this. She first said to evaluate when n = 0, then consider the cases when n = even and n = odd, comparing the even cases to the p-th derivative of Io.

For the Io case, I evaluated it and obtained I_o = \frac{1}{2} \sqrt{\frac{\pi}{\lambda}}

Now, for the "p-th" derivative of Io, i got

\frac{d^p}{d \lambda^2} I_o = \frac{\prod_{p=1}^p (1 - 2p)}{2^{p+1}} \sqrt{\pi} \lambda^{-\frac{(2p + 1)}{2}}

I don't see how this related to n = 2p (even case) where

I2p = \int_0^\infty x^{2p} e^{- \lambda x^2} dx

And even when I do figure this out, does this all combine into one answer, or is it kind of like a piecewise answer?

Any help with what to do with the even/odd cases would be greatly appreciated

Thanks
 
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Hello, I will give some hints here, and probably I will rearrange this quick reply in 24 hours.

\int_0^\infty e^{-ax^2}dx = \frac{1}{2} \sqrt{\frac{\pi}{a}}

\int_0^\infty x e^{-ax^2}dx = \frac{1}{2a}

\int_0^\infty x^n e^{-ax^2}dx = \frac{(n-1)!}{2(2a)^{n/2}} \sqrt{\frac{\pi}{a}}

\int_0^\infty x^{2n+1} e^{-ax^2}dx = \frac{n!}{2 a^{n+1}}

\int_0^\infty x^{2n} e^{-ax^2}dx = \frac{(2n-1)!}{2^{2n+1} a^n} \sqrt{\frac{\pi}{a}}
 

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