7.1.17 int e^{-\dfrac{x^2}{2}} dx from 0 to infty

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In summary, the integral of $\displaystyle\int_0^\infty e^{-\dfrac{x^2}{2}} dx$ can be done in two ways, either by changing the name of the "dummy variable" to I and integrating by dividing by 2, or by using polar coordinates and integrating by rdrd\theta.
  • #1
karush
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Evaluate $\displaystyle\int_0^\infty e^{-\dfrac{x^2}{2}} dx$

ok first reponse is use IBP but can we use $e^u$ where $u=-\dfrac{x^2}{2}$ ot $u=\dfrac{x}{\sqrt{2}}$
 
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  • #2
It's well known that $\displaystyle \begin{align*} \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} = \sqrt{\pi} \end{align*}$, and due to the evenness of this function, that means $\displaystyle \begin{align*} \int_0^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} = \frac{\sqrt{\pi}}{2} \end{align*}$.

If we compare the functions $\displaystyle \begin{align*} f\left( x \right) = \mathrm{e}^{-x^2} \end{align*}$ and $\displaystyle \begin{align*} g\left( x \right) = \mathrm{e}^{-\left( \frac{x}{\sqrt{2}} \right) ^2 } \end{align*}$, we can see that $\displaystyle \begin{align*} g\left( x \right) \end{align*}$ is the image of $\displaystyle \begin{align*} f\left( x \right) \end{align*}$ after a dilation by factor $\displaystyle \begin{align*} \sqrt{2} \end{align*}$ from the $\displaystyle \begin{align*} y \end{align*}$ axis. Therefore, their integrals are also dilated by factor $\displaystyle \begin{align*} \sqrt{2} \end{align*}$.

Therefore $\displaystyle \begin{align*} \int_0^{\infty}{ \mathrm{e}^{-\frac{x^2}{2}}\,\mathrm{d}x } = \sqrt{2}\cdot \frac{\sqrt{\pi}}{2} = \frac{\sqrt{2\,\pi}}{2}\end{align*}$.
 
  • #3
actually i didn't know that...
but sure helps in solving the problem
much Mahalo
 
  • #4
Here is the rather clever way that integral can be done.
Let \(\displaystyle I= \int_0^\infty e^{-\frac{x^2}{2}} dx\).
Then \(\displaystyle I= \int_0^\infty e^{-\frac{y^2}{2}} dy\) since that is just changing the name of the "dummy variable".

So \(\displaystyle I^2= \left( \int_0^\infty e^{-\frac{x^2}{2}} dx\right)\left(\int_0^\infty e^{-\frac{y^2}{2}} dy\right)\)
\(\displaystyle I^2= \int_0^\infty\int_0^\infty e^{-6\frac{x^2+ y^2}{2}} dxdy\).

To integrate that, change to polar coordinates! Since x goes from 0 to infinity and y goes from 0 to infinity, covering the first quadrant, in polar coordinates r goes from 0 to infinity and \(\displaystyle \theta\) goes from 0 to \(\displaystyle \frac{\pi}{2}\). \(\displaystyle x^2+ y^2= r^2\) and \(\displaystyle dxdy= rdrd\theta\).

The integral becomes \(\displaystyle \int_0^{2\pi}\int_0^\infty e^{-\frac{r^2}{2}}rdrd\theta\)
\(\displaystyle = 2\pi\int_0^\infty e^{-\frac{r^2}{2}}rdr\)

Now let \(\displaystyle u= \frac{r^2}{2}\) so that \(\displaystyle du= r dr\) so the integral becomes
\(\displaystyle 2\pi\int_0^\infty e^{-u}du= 2\pi\left[-e^{-u}\right]_0^\infty= 2\pi\left[-(0- 1)\right]= 2\pi\).
 

1. What is the meaning of the numbers and symbols in the expression "7.1.17 int e^{-\dfrac{x^2}{2}} dx from 0 to infty"?

The numbers and symbols in this expression represent a mathematical integral. The "7.1.17" refers to the date (July 1, 2017) and is often used as a label to identify a specific integral problem. The "int" is the symbol for integral, and the "e" represents the mathematical constant e (approximately 2.718). The rest of the expression, e^{-\dfrac{x^2}{2}} dx from 0 to infty, is the integrand (the function being integrated) and the limits of integration, respectively.

2. What is the purpose of "7.1.17 int e^{-\dfrac{x^2}{2}} dx from 0 to infty" in mathematics?

This expression is used to calculate the value of a specific integral. Integrals are important in mathematics because they allow us to find the area under a curve, which has many real-world applications in fields such as physics, engineering, and economics.

3. How do you solve "7.1.17 int e^{-\dfrac{x^2}{2}} dx from 0 to infty"?

To solve this integral, we need to use a mathematical technique called integration. In this specific case, we can use the substitution method, where we substitute a new variable for x and then use the properties of the integral to simplify the expression. After simplifying, we can evaluate the integral using the limits of integration.

4. What is the significance of the limits of integration in "7.1.17 int e^{-\dfrac{x^2}{2}} dx from 0 to infty"?

The limits of integration, 0 and infty, determine the range over which the integral is being evaluated. In this case, the integral is being evaluated from 0 to infinity, meaning that we are finding the area under the curve of the integrand function from 0 to infinity. This is important because it allows us to calculate the total area under the curve, rather than just a specific portion.

5. What are the possible applications of "7.1.17 int e^{-\dfrac{x^2}{2}} dx from 0 to infty"?

This integral has many potential applications in fields such as physics, engineering, and economics. For example, it can be used to calculate the probability of a certain event occurring in a normal distribution, or to determine the amount of work done by a force over a certain distance. It can also be used in financial calculations, such as determining the value of an investment over time.

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