Ok. One more attempt!Solution:Let I=∞∫−∞e−x2dx
Test LaTex 1. I=∞∫−∞e−x2dx
Test with backticks around dollar signs:4. `$I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} $`
Curly brackets test{I=∞∫−∞e−x2dx}
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Curly brackets test {I=∞∫−∞e−x2dx}
Ok. One more attempt!
Solution:
Let I=∞∫−∞e−x2dx
Since x is a dummy variable we can also write: I=∞∫−∞e−y2dy
By multiplying the two together, we get:
I2=∞∫−∞e−x2dx∞∫−∞e−y2dy=∞∫−∞∞∫−∞e−(x2+y2)dxdy
Think of the x and y parameters as Cartesian coordinates, so this double integra;l is the integral over all the plane of
e−(x2+y2)
Change to polar coordinates : x→rcosθ,y→rsinθ,dxdy→rdrdθ
I2=2π∫0dθ∞∫0re−r2dr=2π∞∫0re−r2dr=π∞∫0e−r2dr2=π∞∫0e−udu=π
Therefore, since I2=π
we have I=√π
or ∞∫−∞e−x2dx=√π
Test LaTex
1. I=∞∫−∞e−x2dx
With dollar signs
2. $I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} $
With backticks
3. `I=∞∫−∞e−x2dx
Test with backticks around dollar signs:
ReplyDelete4. `$I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} $`
Curly brackets test
{I=∞∫−∞e−x2dx}
I=∞∫−∞e−x2dx
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ReplyDeleteCurly brackets test {I=∞∫−∞e−x2dx}