- #1
FaradayCage
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Is it possible to integrate [tex]\int \frac{1}{1+e^{-x^2}} dx[/tex]
Solving the Impossible Integral: \int \frac{1}{1+e^{-x^2}} dx
- Thread starter FaradayCage
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Homework Help Overview
The discussion revolves around the integral \(\int \frac{1}{1+e^{-x^2}} dx\) and its potential for integration. Participants explore the nature of this integral within the context of calculus, particularly focusing on whether it can be expressed in terms of elementary functions.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Some participants question the integrability of the function, noting that while it is continuous and defined, its indefinite integral may not be expressible in elementary terms. Others reference related integrals, such as \(\int e^{-x^2} dx\), to draw parallels and highlight challenges in integration.
Discussion Status
The discussion is active, with participants sharing insights and methods related to integration techniques, including a reference to polar coordinates for evaluating specific integrals. There is an acknowledgment of the error function's role in relation to the integral, though no consensus on the original integral's solvability has been reached.
Contextual Notes
Participants express varying levels of familiarity with integration techniques and the implications of using non-elementary functions like the error function. The original poster's inquiry sets the stage for exploring assumptions about integrability and the definitions of related functions.
- #2
tiny-tim
Science Advisor
Homework Helper
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Hi FaradayCage! 
Well, it's not "possible" to integrate the easier indefinite integral ∫ e-x2 dx , so I shouldn't think so.
FaradayCage said:
Well, it's not "possible" to integrate the easier indefinite integral ∫ e-x2 dx , so I shouldn't think so.
- #3
g_edgar
- 606
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Any continuous function is integrable on any bounded interval. So this function is. Its indefinite integral exists. But (as noted) the answer is not an elementary function.
- #4
misnomered
- 1
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Note: This is my first post, so if there is some sort of protocol I'm forgetting or some error in my formatting, go easy for me.
My Calculus III professor showed us a method to calculate [tex]\int_{0}^{\infty} e^{-nx^2}dx[/tex], using polar coordinates.
Think of [tex]x[/tex] and [tex]y[/tex] as independent variables and let
[tex]I = \int_{0}^{\infty} e^{-nx^2}dx = \int_{0}^{\infty} e^{-ny^2}dy.[/tex]
Then
[tex]I^2 = \left(\int_{0}^{\infty} e^{-nx^2}dx\right)\left(\int_{0}^{\infty} e^{-ny^2}dy\right) = \int_{0}^{\infty}\int_{0}^{\infty} e^{-n(x^2 + y^2)}dxdy[/tex]
Substituting into polar coordinates, we have that [tex]r^{2} = x^{2} + y^{2}[/tex] and [tex]dx dy = r dr d\theta[/tex] and thus
[tex]I^2 = \int_{0}^{2\pi}\int_{0}^{\infty} re^{-nr^2}dr d\theta = \left(\int_{0}^{2\pi}d\theta\right)\left(\int_{0}^{\infty} re^{-nr^2}dr\right) = 2\pi\int_{0}^{\infty} re^{-nr^2}dr[/tex].
We can then take [tex]u = r^2[/tex] so that [tex]\frac{du}{2} = r dr[/tex] and thus
[tex] I^{2} = 2\pi\int_{0}^{\infty} re^{-nr^2}dr = \pi\int_{0}^{\infty} e^{-nu}du = \frac{\pi}{n}[/tex]
and thus
[tex] I = \sqrt{\frac{\pi}{n}}.[/tex]
If you change the bounds of the integral however, calculating the integral becomes very hard.
My Calculus III professor showed us a method to calculate [tex]\int_{0}^{\infty} e^{-nx^2}dx[/tex], using polar coordinates.
Think of [tex]x[/tex] and [tex]y[/tex] as independent variables and let
[tex]I = \int_{0}^{\infty} e^{-nx^2}dx = \int_{0}^{\infty} e^{-ny^2}dy.[/tex]
Then
[tex]I^2 = \left(\int_{0}^{\infty} e^{-nx^2}dx\right)\left(\int_{0}^{\infty} e^{-ny^2}dy\right) = \int_{0}^{\infty}\int_{0}^{\infty} e^{-n(x^2 + y^2)}dxdy[/tex]
Substituting into polar coordinates, we have that [tex]r^{2} = x^{2} + y^{2}[/tex] and [tex]dx dy = r dr d\theta[/tex] and thus
[tex]I^2 = \int_{0}^{2\pi}\int_{0}^{\infty} re^{-nr^2}dr d\theta = \left(\int_{0}^{2\pi}d\theta\right)\left(\int_{0}^{\infty} re^{-nr^2}dr\right) = 2\pi\int_{0}^{\infty} re^{-nr^2}dr[/tex].
We can then take [tex]u = r^2[/tex] so that [tex]\frac{du}{2} = r dr[/tex] and thus
[tex] I^{2} = 2\pi\int_{0}^{\infty} re^{-nr^2}dr = \pi\int_{0}^{\infty} e^{-nu}du = \frac{\pi}{n}[/tex]
and thus
[tex] I = \sqrt{\frac{\pi}{n}}.[/tex]
If you change the bounds of the integral however, calculating the integral becomes very hard.
- #5
sagardip
- 38
- 0
What i think is that ∫ e-x2 dx is a reduced erf(x) function or error function.
erf(x)=2/(pi^0.5) ∫ e-x2 dx
erf(x)=2/(pi^0.5) ∫ e-x2 dx
- #6
Gib Z
Homework Helper
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You are correct, but the error function is simply what we define to be the integral it is, well, defined by. It's not some sort of theorem or result, the error function is not some different type of function that happens to be equal to this integral.