Help with a definite integral property

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Discussion Overview

The discussion revolves around the properties of definite integrals, specifically focusing on the integral of the function \( e^{x^2} \) over the entire real line and its relationship to integrals over positive and negative domains. Participants explore the implications of the function being even and the conditions under which the limits of integration can be altered.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Luc questions the validity of changing the limits of integration from \(-\infty\) to \(0\) and \(0\) to \(+\infty\) for the integral of \( e^{x^2} \).
  • Some participants point out that \( e^{x^2} \) is an even function, which leads to symmetry about the y-axis, allowing the integral from \(-\infty\) to \(+\infty\) to be expressed as twice the integral from \(0\) to \(+\infty\).
  • Luca expresses confusion about the limits in the context of the symmetry argument and whether the limits can indeed be changed correctly.
  • Another participant notes that all integrals diverge unless the integrand is modified (e.g., using \( e^{-x^2} \)).
  • There is a discussion about the importance of checking for convergence or divergence in integrals.
  • Some participants affirm that the change of limits for the integral still holds due to the properties of even functions.

Areas of Agreement / Disagreement

Participants generally agree on the symmetry of the function \( e^{x^2} \) and its implications for changing limits of integration. However, there is disagreement regarding the convergence of the integrals, with some noting that the integrals diverge unless modified.

Contextual Notes

There are unresolved issues regarding the convergence of the integrals discussed, and the implications of changing limits depend on the nature of the function being integrated.

pamparana
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Hello,

Saw this expression:

[tex]^{inf}_{-inf} \int e^{x^{2}} = 2 \times ^{inf}_{+0} \int e^{x^{2}}[/tex]

I am unable to understand this change of base. You can change from -inf to 0 and 0 to +inf but do not see how that equals to 2 * the integral from 0 to infinity.

I hope someone will help me see the light :)

Thanks,

Luc
 
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it's because [tex]e^{x^2}[/tex] is an even function, that is, it satisfies the equation f(-x) = f(x). if you look at the graph of the function it's symmetric about the y-axis
 
That makes sense. Thanks for that. I still have one confusion though:

[tex]\int^{\infty}_{-\infty} e^{x^{2}} = \int^{0}_{-\infty} e^{x^{2}} + \int^{\infty}_{0} e^{x^{2}}[/tex]

Now using the function symmetry we have:

[tex]\int^{\infty}_{-\infty} e^{x^{2}} = \int^{-\infty}_{0} e^{x^{2}} + \int^{\infty}_{0} e^{x^{2}}[/tex]

However, the limits are still messed up in the first term, right? The upper limit still has [tex]-\infty[/tex]

Or have I done this completely wrong?

Thanks,

Luca
 
[itex]\int_{-\infty}^{\infty} e^{x^2} dx = \int_{-\infty}^{0} e^{x^2} dx + \int_{0}^{\infty} e^{x^2} dx[/itex] but since [itex]e^{x^2}[/itex] is even, [itex]\int_{-\infty}^{0} e^{x^2} dx = \int_{0}^{\infty} e^{x^2} dx[/itex] so [itex]\int_{-\infty}^{\infty} e^{x^2} dx = 2\int_{0}^{\infty} e^{x^2} dx[/itex]
 
Ahhhh thanks. Did not know you could change the integral limits for an even function like that.

Cheers,
Luca
 
And, of course, all the integrals mentioned above diverge unless the integrand has a negative sign:

[tex]e^{-x^2}[/tex]

:rolleyes:
 
ARGHHHH i don't know how i missed that! i guess it's too late to delete everything I've added to this thread :mad: luckily for me math isn't an exact science
 
Last edited:
Of course. Checking for convergence or divergence is another business all-together.

However, the change of limits for the integral still holds, right?

Thanks,
Luca
 
yes, because [itex]\int_{0}^{\infty} f(x) dx = \lim_{M \rightarrow \infty}\int_{0}^{M} f(x) dx[/itex]
 

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