- #1

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## Main Question or Discussion Point

**Integrating exp(x^2) like gaussian integral??**

Hi,

I can't solve this integral [tex]\int^{1}_{0}\\e^{x^2}\\dx[/tex]

Can I solve this integral like gaussian integral?

Please help me

Thanks.

- Thread starter coki2000
- Start date

- #1

- 91

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Hi,

I can't solve this integral [tex]\int^{1}_{0}\\e^{x^2}\\dx[/tex]

Can I solve this integral like gaussian integral?

Please help me

Thanks.

- #2

- 586

- 1

The Gaussian integral you are referring to is usually being integrated from minus infinity to plus infinity. Integrals like yours can be expressed in terms of the so-called error function but not in terms of more elementary functions.

The numerical value of the integral in your case is 1.46265.

- #3

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I found a solvation but I am not sure that the solvation is right.Firstly I write the integral in cartesian coordinates as surface integral after in polar coordinates.

Where [tex]r=\sqrt{x^2+y^2}[/tex]

[tex]\int_S {e^{r^2}}dA=\int_{0}^{1} \int_{0}^{1} e^{{\sqrt{x^2+y^2}}^2}dxdy=\int_{0}^{1} e^{x^2}dx\int_{0}^{1} e^{y^2}dy=[\int_{0}^{1}e^{x^2}]^2[/tex]

[tex]\int_S {e^{r^2}}dA=\int_{0}^{\pi /4} \int_{0}^{1} e^{r^2}rdrd\theta=\frac{\pi }{8}{(e-1)}=[\int_{0}^{1}e^{x^2}]^2[/tex]

I determine the limits from graph of [tex]e^{x^2}[/tex]

[tex]\int_{0}^{1}e^{x^2}=\sqrt{\frac{\pi }{8}{(e-1)}}[/tex]

I solve this problem like solvation of gaussian integral but I'm in doubt that it is right.

Please help me and show me right solvation.

Thanks.

Last edited:

- #4

- 586

- 1

Your result is incorrect and the reason is that the rectangle [0,1] x [0,1] in Cartesian coordinates cannot easily be parametrized in polar coordinates, in particular it does not correspond to [0,pi/4] x [0,1] , which describes an 45 degree sector of the unit disc.

- #5

Gib Z

Homework Helper

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If you don't mind getting a infinite series for an answer, replacing the integrand with its Taylor series and integrating term by term gives a relatively nice looking series:

[tex]\sum_{n=0}^{\infty} \frac{1}{(2n+1) n!}[/tex]

- #6

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If that is so, then what can we do to solve this integral?I worked to solve this integral like the solvation of gaussian integral but I can't find the parameters of integrals.If this solution is right, what is the parameters of integrals else what is the right solution of this problem?

I really wonder the solution.

Please help me,

Thanks in advance.

- #7

Gib Z

Homework Helper

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It has no more elementary expression other than in terms of the Error Function, which is pretty much just shorthand notation for that integral anyway, or you could try the series I put up if you like that better. There is NO "nicer" solution in terms of things like pi and e.

- #8

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I gave this series to Maple, and the result it gave me involves the error function in the form erfc:

[tex]\sum_{n=0}^{\infty} \frac{1}{(2n+1) n!}[/tex]

[tex]\frac{i\sqrt {\pi }}{2} \left( -1+{\rm erfc} \left( i \right) \right) [/tex]

- #9

Gib Z

Homework Helper

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Erfc is the complementary function of the error function, and is often defined to be 1-erf(x).I gave this series to Maple, and the result it gave me involves the error function in the form erfc:

[tex]\frac{i\sqrt {\pi }}{2} \left( -1+{\rm erfc} \left( i \right) \right) [/tex]

Since [tex]erf(x)=\frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} dt[/tex] and through some formal manipulations, we first let x= i, and then change variables via u= it we can easily show how the expression maple gives is indeed [tex]\int^1_0 e^{t^2} dt[/tex]

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