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How does one calculate the indefinite integral? $$\int\frac1{\sigma\sqrt{2\pi}}\exp\left(-\frac{x^2}{2\sigma^2}\right)dx$$ Where $\sigma$ is some constant.


Work so far:

Integrating from exp as rest is constant. $$\begin{align} \int\exp\left(-\frac{x^2}{2\sigma^2}\right)dx&=\sum_{n=0}^\infty\frac{\left(-\frac{x^2}{2\sigma^2}\right)^n}{n!}=-\sum_{n=0}^\infty n!^{-1}2^{-1}\sigma^{-2n}\int x^{2n}dx\\ &=-\sum_{n=0}^\infty n!^{-1}\sigma^{-2n}x^{2n}x^{-1}\\ \end{align}$$

I pulled it apart, integrated it, now I cant put it back together.

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  • $\begingroup$ I am aware of other attempts of doing the integration but they are all definite. $\endgroup$ – Ali Caglayan Aug 6 '13 at 22:04
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In general, the integral

$$\int e^{-x^2} dx$$

cannot be expressed in terms of elementary functions. For a particular definite integral, we can define the error function,

$$\text{erf }{x} = \frac{2}{\pi} \int_0^x e^{-x^2} dx$$

In order to introduce constants as in your function, a simple substitution and rescaling can be done.

On the other hand, if you want to compute the number

$$\int_{\mathbb{R}} e^{-x^2} dx$$

the usual trick is to square the integral, convert into polar coordinates, and evaluate.

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It's been a while since you've asked, but this is the indefinite integral:

[\begin{array}{l} \int {\left( {\frac{{{e^{ - \frac{1}{2}\;\cdot\;{{\left( {\frac{x}{\sigma }} \right)}^2}}}}}{{\sigma \sqrt {2\pi } }}} \right)\;dx = } \frac{{\int {\left( {\frac{{{e^{ - \frac{1}{2}\;\cdot\;{{\left( {\frac{x}{\sigma }} \right)}^2}}}}}{\sigma }} \right)} \;dx}}{{\sqrt {2\pi } }}\\ u = \frac{x}{\sigma }\\ du = \frac{1}{\sigma }\;dx\\ = \frac{{\int {\left( {{e^{ - \frac{{{u^2}}}{2}}}} \right)} \;du}}{{\sqrt {2\pi } }}\\ = \frac{{\int {{e^{ - \frac{{{u^2}}}{2}}}} \;du}}{{\sqrt {2\pi } }}\\ = \frac{{\int {{e^{ - \frac{{{u^2}}}{2}}}} \cdot1\;du}}{{\sqrt {2\pi } }}\\ = \frac{{{e^{ - {u^2}}}\cdot\int 1 \;du - \int {\left( { - u{e^{ - \frac{{{u^2}}}{2}}}\cdot\int 1 \;du} \right)} \;du}}{{\sqrt {2\pi } }}\\ = \frac{{\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1} - \int {\left( {\frac{{ - {u^2}{e^{ - \frac{{{u^2}}}{2}}}}}{1}} \right)\;} \;du}}{{\sqrt {2\pi } }}\\ = \frac{{\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1}\; - \frac{{ - 2{e^{ - \frac{{{u^2}}}{2}}}}}{1}\;\cdot\;\int {{u^2}} \;du + \int {\left( {\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1}\;\cdot\int {{u^2}} \;du} \right)} \;du}}{{\sqrt {2\pi } }}\\ = \frac{{\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1}\; + \frac{{{u^3}{e^{ - \frac{{{u^2}}}{2}}}}}{3} + \int {\left( {\frac{{{u^4}{e^{ - \frac{{{u^2}}}{2}}}}}{3}} \right)} \;du}}{{\sqrt {2\pi } }}\\ = \frac{{\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1}\; + \frac{{{u^3}{e^{ - \frac{{{u^2}}}{2}}}}}{3} + \frac{{{e^{ - \frac{{{u^2}}}{2}}}}}{3}\;\cdot\int {\left( {{u^4}} \right)} \;du - \int {\left( {\frac{{ - u{e^{ - \frac{{{u^2}}}{2}}}}}{3}\;\cdot\int {{u^4}} \;du} \right)} \;du}}{{\sqrt {2\pi } }}\\ = \frac{{\frac{{u{e^{ - \frac{{{u^2}}}{2}}}}}{1}\; + \frac{{{u^3}{e^{ - \frac{{{u^2}}}{2}}}}}{3} + \frac{{{u^5}{e^{ - \frac{{{u^2}}}{2}}}}}{{15}} - \int {\left( {\frac{{ - {u^6}{e^{ - \frac{{{u^2}}}{2}}}}}{{15}}} \right)} \;du}}{{\sqrt {2\pi } }}\\ = \frac{{\sum\limits_{n = 1}^\infty {\left( {\frac{{{u^{2n - 1}}\cdot{e^{ - \frac{{{u^2}}}{2}}}}}{{\prod\limits_{i = 1}^n {\left( {2i - 1} \right)} }}} \right)} }}{{\sqrt {2\pi } }}\\ = \sum\limits_{n = 1}^\infty {\left( {\frac{{{{\left( {\frac{x}{\sigma }} \right)}^{2n - 1}}\cdot{e^{ - \frac{1}{2}\;\cdot\;{{\left( {\frac{x}{\sigma }} \right)}^2}}}}}{{\sqrt {2\pi } \;\cdot\prod\limits_{i = 1}^n {\left( {2i - 1} \right)} }}} \right)} \end{array}]

$$ \int \frac{e^{-\frac{x^2}{2\sigma^2}}}{\sigma \sqrt{2\pi}} ~dx = \sum_{n=1}^\infty \left( \frac{\left(\frac{x^2}{\sigma^2}\right)^{2n-1} e^{-\frac{x^2}{2\sigma^2}}}{\sqrt{2\pi}\cdot \prod_{i=1}^n (2i-1)} \right) $$

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  • $\begingroup$ We can use U substitution, u=x/σ, so then du=1/σ dx. The sigma in the denominator on the left comes from the chain rule. We then shift the 1/√(2π) out as it is a constant coefficient. we are left with ∫ e^-x^2 dx, we can say this is multiplied by 1, and use integration by parts repeatedly. Then, to check if you did get the correct integral, you can differentiate it, and check if it is the same as your integrand. $\endgroup$ – user123850 Jan 26 '14 at 17:01
  • $\begingroup$ How about this... I'll edit your Answer, giving the MathJax/LaTeX equivalent of your image, and give you some ideas about how to go about including the steps you describe above into your Answer. I'll also reset my downvote, so you have a fresh start. $\endgroup$ – hardmath Jan 26 '14 at 17:06

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