Integrate int e^{-u^2} du.



{eq}\int e^{-u^2} du {/eq}.

Integrals without Antiderivatives

If we are able to find the antiderivative of an integrand, the Fundamental Theorem of Calculus allows us to define this integral as a new function. However, if we cannot find the antiderivative of an integrand, then we cannot evaluate the integral in this manner. If this is a definite integral, we need to use numerical approximations to find the value. For an indefinite integral, we need to look to known definitions.

Answer and Explanation:

This integrand doesn't have a known antiderivative. In fact, this is a famous integral, called the Gaussian integral, and it's incredibly useful in statistics. This is because this function is related to the normal distribution.

Since this function doesn't have a defined antiderivative, we need to apply definitions to find the function that this integral represents. However, this definition requires us to define a new function, called the error function. This function is defined using this integral, so finding another way to express this function so that we can interpret it is incredibly difficult.

{eq}\displaystyle \text{erf}(x) = \frac{2}{\sqrt {\pi}} \int_0^x e^{-u^2}{du} {/eq}

With this definition given, we can solve for the integral to find that it has the following function representation.

{eq}\displaystyle \int e^{-u^2} du = \frac{\sqrt{\pi}}{2} \text{erf}(x) + c {/eq}

If we didn't mind finding an approximation, we could find the Maclaurin series for this integrand and then integrate it. This Maclaurin series can be found by evaluating the standard Maclaurin series for {eq}e^x {/eq} at {eq}-u^2 {/eq}:

{eq}\begin{align*} e^x& \approx 1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!} x^4 + ...\\ e^{-u^2}&\approx 1 + (-u^2) + \frac{1}{2}(-u^2)^2 + \frac{1}{3!}(-u^2)^3 + \frac{1}{4!} x(-u^2)^4 + ...\\ &\approx 1 - u^2 + \frac{1}{2}u^4 - \frac{1}{3!}u^6 + \frac{1}{4!}u^8 +... \end{align*} {/eq}

Integrating this will give a polynomial approximation for this function near zero.

{eq}\begin{align*} \int e^{-u^2}&\approx \int (1 - u^2 + \frac{1}{2}u^4 - \frac{1}{3!}u^6 + \frac{1}{4!}u^8+...) du\\ &= u - \frac{1}{3}u^3 + \frac{1}{10}u^5 - \frac{1}{42}u^7 + \frac{1}{216}u^9+...+c \end{align*} {/eq}

Learn more about this topic:

Indefinite Integrals as Anti Derivatives

from Math 104: Calculus

Chapter 12 / Lesson 11

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