Question

What is the integral of the error function?

Answer 1

To solve this question we need to know the concept of integration. The error function is denoted by erf, which defined as $\text{erf z = }\dfrac{2}{\sqrt{\pi }}\int\limits_{0}^{z}{{{e}^{-{{t}^{2}}}}dt}$. To integrate the error function we will firstly integrate the term by using integration by parts. The second step will be to solve the problem and will be to use a substitution method of integration.

Complete step by step answer:
The question ask us to find the integral of the error function which is $\text{erf z = }\dfrac{2}{\sqrt{\pi }}\int\limits_{0}^{z}{{{e}^{-{{t}^{2}}}}dt}$. $\int{\dfrac{2}{\sqrt{\pi }}}z{{e}^{-{{z}^{2}}}}dz$We will first integrate the function by the method of integration by parts:
Let us consider the function $u$ as $\text{erf z}$ and $dv$ as $dt$ which means:
$\Rightarrow \text{u = erf z}$ and $dv=dt$
We will now use the fundamental theorem of calculus, on using this we get:
$du=\dfrac{2}{\sqrt{\pi }}{{e}^{-{{z}^{2}}}}$ and $v=z$
By using the formula for the integration by parts formula we get:
$\int{udv=uv-\int{vdu}}$
On substituting the value of the $u$and $v$ we get the following equation which is given below:
$\int{\text{erf }}\left( z \right)dz=zerf\left( z \right)-\int{\dfrac{2}{\sqrt{\pi }}}z{{e}^{-{{z}^{2}}}}dz$
The next step to evaluate the problem is to calculate it by the integration by substitution:
To apply the formula for the integration by substitution to solve the equation$\int{\dfrac{2}{\sqrt{\pi }}}z{{e}^{-{{z}^{2}}}}dz$, for doing this let us consider $u=-{{z}^{2}}$ , so on differentiating it we get:
$\Rightarrow du=-2zdz$ and so
$\Rightarrow -\int{\dfrac{2}{\sqrt{\pi }}z{{e}^{-{{z}^{2}}}}dz=\dfrac{1}{\sqrt{\pi }}\int{{{e}^{u}}du}}$
$\Rightarrow \dfrac{1}{\sqrt{\pi }}{{e}^{u}}+C$
$\Rightarrow \dfrac{{{e}^{-{{z}^{2}}}}}{\sqrt{\pi }}+C$
Putting it all together we get the final result which is:
$\int{\text{erf }\left( z \right)dz=z\text{erf}\left( \text{z} \right)}+\dfrac{{{e}^{-{{z}^{2}}}}}{\sqrt{\pi }}+C$

Note: There are questions in which integration becomes difficult so we need to expand the function to solve the equation individually so that there is no scope of error. In a single question there tend to be situations where more than one property of integration is used to solve the question, as could be seen in the above question.