Ok. One more attempt!Solution:Let \[I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \]Since x is a dummy variable we can also write: \[I = \int\limits_{ - \infty }^\infty {{e^{ - {y^2}}}dy} \] By multiplying the two together, we get: \[\begin{array}{c}{I^2} = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \int\limits_{ - \infty }^\infty {{e^{ - {y^2}}}dy} \\ = \int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {{e^{ - ({x^2} + {y^2})}}} dxdy} \end{array}\]Think of the x and y parameters as Cartesian coordinates, so this double integra;l is the integral over all the plane of \[{e^{ - ({x^2} + {y^2})}}\]Change to polar coordinates : \[x \to r\cos \theta ,\quad y \to r\sin \theta ,\quad dxdy \to rdrd\theta \] to get: \[\begin{array}{c}{I^2} = \int\limits_0^{2\pi } {d\theta \int\limits_0^\infty {r{e^{ - {r^2}}}dr} } \\ = 2\pi \int\limits_0^\infty {r{e^{ - {r^2}}}dr} \\ = \pi \int\limits_0^\infty {{e^{ - {r^2}}}d{r^2}} \\ = \pi \int\limits_0^\infty {{e^{ - u}}du} \\ = \pi \end{array}\]Therefore, since \[{I^2} = \pi \] we have \[I = \sqrt \pi \] or \[\int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} = \sqrt \pi \]
Test LaTex 1. \[I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \]With dollar signs2. $I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} $With backticks3. `\[I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \]`
Test with backticks around dollar signs:4. `$I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} $`
Curly brackets test\[ \{I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \} \]\[ \\{I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \\} \]
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Curly brackets test \[ \{I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \} \] \[ \\{I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \\} \] on
Ok. One more attempt!
ReplyDeleteSolution:
Let \[I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \]
Since x is a dummy variable we can also write: \[I = \int\limits_{ - \infty }^\infty {{e^{ - {y^2}}}dy} \]
By multiplying the two together, we get:
\[\begin{array}{c}
{I^2} = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \int\limits_{ - \infty }^\infty {{e^{ - {y^2}}}dy} \\
= \int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {{e^{ - ({x^2} + {y^2})}}} dxdy}
\end{array}\]
Think of the x and y parameters as Cartesian coordinates, so this double integra;l is the integral over all the plane of
\[{e^{ - ({x^2} + {y^2})}}\]
Change to polar coordinates : \[x \to r\cos \theta ,\quad y \to r\sin \theta ,\quad dxdy \to rdrd\theta \] to get:
\[\begin{array}{c}
{I^2} = \int\limits_0^{2\pi } {d\theta \int\limits_0^\infty {r{e^{ - {r^2}}}dr} } \\
= 2\pi \int\limits_0^\infty {r{e^{ - {r^2}}}dr} \\
= \pi \int\limits_0^\infty {{e^{ - {r^2}}}d{r^2}} \\
= \pi \int\limits_0^\infty {{e^{ - u}}du} \\
= \pi
\end{array}\]
Therefore, since \[{I^2} = \pi \]
we have \[I = \sqrt \pi \]
or \[\int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} = \sqrt \pi \]
Test LaTex
ReplyDelete1. \[I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \]
With dollar signs
2. $I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} $
With backticks
3. `\[I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \]`
Test with backticks around dollar signs:
ReplyDelete4. `$I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} $`
Curly brackets test
ReplyDelete\[ \{I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \} \]
\[ \\{I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \\} \]
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ReplyDeleteCurly brackets test \[ \{I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \} \] \[ \\{I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} \\} \] on
ReplyDelete