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Antiderivative of $x{e}^{-c{x}^{2}}$
I need to define c in ${\int }_{0}^{\mathrm{\infty }}x{e}^{-c{x}^{2}},$, so that it becomes a probability-mass function (so that it equals 1).
Where do I even begin finding the antiderivative of this? I know the answer will be: $\frac{{e}^{-c{x}^{2}}}{-2c}$.
Trying to use partial integration:
$\int f\left(x\right)g\left(x\right)dx=f\left(x\right)G\left(x\right)-\int {f}^{\prime }\left(x\right)G\left(X\right)$
and picking x as my g(x), and ${e}^{-c{x}^{2}}$ as my f(x) I end up with:
${f}^{\prime }\left(x\right)=-2cxf\left(x\right)$
$G\left(X\right)=\frac{{x}^{2}}{2}$
${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int }_{0}^{\mathrm{\infty }}-2cxf\left(x\right)\frac{{x}^{2}}{2}dx$
Which simplifies to: ${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int }_{0}^{\mathrm{\infty }}-c{x}^{4}{e}^{-c{x}^{2}}dx$, which I find is just a mess.
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Amir Beck
Step 1
The antiderivative is easy. Note that ,since you want the integral to equal 1:
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}x{e}^{-c{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=1& ⇔\frac{1}{2c}{\int }_{0}^{\mathrm{\infty }}2cx{e}^{-c{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=1\\ & ⇔\frac{1}{2c}{\left[-{e}^{-c{x}^{2}}\right]}_{0}^{\mathrm{\infty }}=1\\ & ⇔2c=1\\ & ⇔c=1/2\end{array}$
Note also that c must be positive , otherwise the integral won't converge.
Step 2
Update: I fixed a major blunder. The quantity in brackets is evaluated as follows:
${\int }_{0}^{\mathrm{\infty }}2cx{e}^{-c{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx=-\underset{x\to +\mathrm{\infty }}{lim}{e}^{-c{x}^{2}}+{e}^{0}=1$
(which is the definition of improper integral) and hence we go on to solve the equation.
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glitinosim3
Explanation:
For $f\left(x\right)={e}^{-c{x}^{2}}$ you have ${f}^{\prime }\left(x\right)=-2cxf\left(x\right)$, so integration by parts gives $\int xf\left(x\right)dx=\frac{1}{-2c}\int {f}^{\prime }\left(x\right)dx=\frac{1}{-2c}f\left(x\right)+C$.
Now the definite integral.
To have a probability-mass function, c has to be positive (otherwise the limit in infinity is intinite) and then $1=\int xf\left(x\right)dx=\frac{1}{-2c}f\left(x\right){|}_{0}^{\mathrm{\infty }}=\frac{1}{-2c}\left(0-1\right)$, so $c=\frac{1}{2}$ is the only solution.