Calculul integralei pe R a functiei e^{-x^2}
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- Ovidiu Jianu
- Posts: 2
- Joined: Thu Sep 27, 2007 8:31 pm
- Location: Bucuresti
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Calculul integralei pe R a functiei e^{-x^2}
Dati cat mai multe solutii distincte pentru calculul integralei \( \int_{-\infty}^{\infty}e^{-x^2}dx \).
Last edited by Ovidiu Jianu on Fri Oct 05, 2007 5:47 pm, edited 1 time in total.
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Re: integrala din e la x patrat dx
I don't speak romanian, but I think you may understand me.jianuovidiu wrote:\( \int_{-\infty}^{\infty}e^{-x^2}dx \).
\( \begin{eqnarray*}
I &=& \int_{ - \infty }^\infty {e^{ - x^2 } \,dx}\\
I^2 &=& \int_{ - \infty }^\infty {e^{ - x^2 } \,dx} \, \cdot \int_{ - \infty }^\infty {e^{ - y^2 } \,dy}\\
&=& \int_{ - \infty }^\infty {\int_{ - \infty }^\infty {e^{ - \left( {x^2 + x^2 } \right)} \,dx\,dy} }
\end{eqnarray*} \)
Introducing polar coordinates defined by \( x=r\cos\theta \) & \( y=r\sin\theta \), it remains to compute \( I^2 = \int_0^\infty {\int_0^{2\pi } {re^{ - r^2 } \,dr\,d\theta } } \).
It follows that \( I^2=\pi \) and we happily get \( I=\sqrt{\pi} \).