How to Integrate x^2 e^(-|x|/b) from -Infinity to Infinity?

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Discussion Overview

The discussion revolves around the evaluation of the integral \(\int_{-\infty}^{\infty} x^{2} e^{-\frac{|x|}{b}} dx\). Participants explore various methods for solving this integral, including substitution and integration by parts, while addressing the implications of the parameter \(b\) on the solution.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes splitting the integral into two parts, from 0 to \(\infty\) and from \(-\infty\) to 0, suggesting the answer might be \(2b\), but expresses uncertainty about its correctness.
  • Another participant agrees that splitting the integral is a valid approach but challenges the assertion that the answer does not involve \(\pi\).
  • A different participant notes that the integrand is even, allowing for simplification to \(2\int_{0}^{\infty} x^{2} e^{-\frac{x}{b}} dx\).
  • One suggestion involves a substitution \(t = -x/b\) and performing integration by parts twice.
  • Another participant suggests a simpler substitution \(t = x/b\) to avoid changing limits.
  • One participant mentions that there are two answers due to the unknown value of \(b\).
  • Another proposes substituting \(b \rightarrow 1/p\) to facilitate integration and differentiating with respect to \(p\) to derive a factor of \(x^2\) in the integrand, ultimately suggesting the answer is \(4b^3\).
  • A participant expresses confusion about the method involving differentiation under the integral sign and requests a full solution.
  • One participant provides a link to a detailed explanation of the differentiation under the integral sign technique.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct method or final answer for the integral. Multiple competing views and methods are presented, indicating ongoing debate and uncertainty.

Contextual Notes

Some participants note that if \(b\) is negative, the integral diverges, highlighting a limitation in the assumptions made during the discussion.

Morto
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Hello!

I need to calculate [tex]\int_{-\infty}^{\infty} x^{2} e^{-\frac{|x|}{b} dx[/tex] and I know the answer should involve [tex]\pi[/tex].
I've tried splitting the integral for x and -x with limits from 0 to infty and -infinty to 0, which gives me the answer of 2b. But I don't think this is correct. Does anyone have any hints or tips on what to do?
 
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Morto said:
I've tried splitting the integral for x and -x with limits from 0 to infty and -infinty to 0,
That's a correct way to approach this.

which gives me the answer of 2b. But I don't think this is correct. Does anyone have any hints or tips on what to do?

Post your work here, and we'll try to help you.

Oh, BTW, the answer does not involve pi, unless you want to write it as answer*pi/pi or something. :wink:
 
Note that your integrand is even. Thus, you may simplify your problem as follows:
[tex]\int_{-\infty}^{\infty}x^{2}e^{-\frac{|x|}{b}}dx=2\int_{0}^{\infty}x^{2}e^{-\frac{x}{b}}[/tex]
where I've omitted the absolute value sign about x, since for x>=0, we have |x|=x.
 
Here is a continuation of arildno post. Let [tex]t=-x/b[/tex] (assuming [tex]b>0[/tex]) and preform substitution. Then do integration by parts twice.
 
Wouldn't it just be simpler to let t=x/b so that we don't need to change our limits to 0 and and negative infinity?
 
Why do a normal substitution?

It looks like one should use integration by parts twice.
 
One more hint, there are two answers. The reason is that you don't know what the value of b is.
 
You don't need to do integration by parts. Just substitute b ---> 1/p
so that the exponential becomes exp(-p x) . Integrate exp(p x) and differentiate twice w.r.t. p to bring down a factor x^2 in the integrand.

The integral is thus 1/p. Differentiating twice w.r.t. p gives 2/p^3. Multiply by 2 and put back b to obtain the answer: 4 b^3.
 
I don't understand what method you are using. Is this some sort of DE trick? I would appreciate learning this method. Please post the full solution.

(Note: If b is negative, the integral diverges.)
 
Last edited:
  • #10
It is explained in detail http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign" .

E.g.

[tex]\int_{0}^{\infty}\exp(-p x)dx=\frac{1}{p}[/tex]

Differentiate both sides w.r.t. p:

[tex]\frac{d}{dp}\int_{0}^{\infty}\exp(-p x)dx=-\frac{1}{p^2}[/tex]

Interchange the order of integration and differentiation to obtain:

[tex]\int_{0}^{\infty}x\exp(-p x)dx=\frac{1}{p^2}[/tex]
 
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