Solving Improper Integrals with Integration by Parts

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Homework Help Overview

The problem involves calculating the improper integral \(\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}x^{2}e^{-\frac{x^{2}}{2}}dx\), utilizing the known result that \(\int^{\infty}_{-\infty}e^{-\frac{x^{2}}{2}}dx=\sqrt{2\pi}\). The original poster suggests using integration by parts as a method to approach the solution.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts integration by parts but struggles to find a suitable combination of \(u\) and \(dv\) that leads to a manageable integral. Some participants suggest that a careful choice of \(dv\) could simplify the process. Others inquire about the specific steps taken by the original poster to identify any mistakes.

Discussion Status

The discussion is ongoing, with participants providing hints and encouraging the original poster to share their work for further analysis. There is a recognition of the potential for integration by parts to be a viable approach, though the original poster expresses difficulty in aligning their results with known outcomes.

Contextual Notes

Some participants note that the integrand represents the density of the standard normal distribution, which may influence the interpretation of the integral's value. The original poster has also indicated that their calculations led to a divergent result, prompting further examination of their approach.

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Homework Statement



Calculate [tex]\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}x^{2}e^{-\frac{x^{2}}{2}}dx[/tex]

Use the fact that [tex]\int^{\infty}_{-\infty}e^{-\frac{x^{2}}{2}}dx=\sqrt{2\pi}[/tex]

Homework Equations



I'm assuming that integration by parts is the best way to solve this.

http://www.math.hmc.edu/calculus/tutorials/int_by_parts/"

The Attempt at a Solution



I want to use integration by parts in order to solve this. I've attempted both combinations of u and v'. However, I'm not able to get an integral in any of my solutions that looks like the one above that is set equal to [tex]\sqrt{2\pi}[/tex].

I have plugged the equation into Mathematica and the answer comes out to be 1. However, getting my work to back that up is proving to be difficult.

Is integration by parts the right way to go about solving this one? Any help is greatly appreciated.
 
Last edited by a moderator:
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The integrand is the density of the standard normal distribution, which is why Mathematica gave you the answer it did.

You should try integration by parts; with a judicious choice of [tex]dv[/tex] you will find that the

[tex] \int \, dv[/tex]

portion (to calculate [tex]v[/tex]) is easily done. After the step, and you have

[tex] \int u \dv = uv - \int v \, du[/tex]

you will see the reason for the hint.
 
UMich1344 said:

Homework Statement



Calculate [tex]\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}x^{2}e^{-\frac{x^{2}}{2}}dx[/tex]

Use the fact that [tex]\int^{\infty}_{-\infty}e^{-\frac{x^{2}}{2}}dx=\sqrt{2\pi}[/tex]

Homework Equations



I'm assuming that integration by parts is the best way to solve this.

http://www.math.hmc.edu/calculus/tutorials/int_by_parts/"
Yes, we know what integration by parts is! How about showing exactly what you did so we can point out any mistakes?

The Attempt at a Solution



I want to use integration by parts in order to solve this. I've attempted both combinations of u and v'. However, I'm not able to get an integral in any of my solutions that looks like the one above that is set equal to [tex]\sqrt{2\pi}[/tex].

I have plugged the equation into Mathematica and the answer comes out to be 1. However, getting my work to back that up is proving to be difficult.

Is integration by parts the right way to go about solving this one? Any help is greatly appreciated.
 
Last edited by a moderator:
I have attached my work with this post. I felt that I was heading in the right direction, but obviously I must have done something wrong near the beginning I would assume, since by the end I found that my integral diverged.
 

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