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

• MHB
• karush
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.
karush
Gold Member
MHB
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 $u=\dfrac{x}{\sqrt{2}}$

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*}.

actually i didn't know that...
but sure helps in solving the problem
much Mahalo

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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