Question

What is the integral of the error function?

Answer 1

Hint: To solve this question we need to know the concept of integration. The error function is denoted by erf, which defined as $\text{erf z = }{\displaystyle \frac{2}{\sqrt{\pi }}}\underset{0}{\overset{z}{\int }}{e}^{-t^2}dt$. 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 $\text{erf z = }{\displaystyle \frac{2}{\sqrt{\pi }}}\underset{0}{\overset{z}{\int }}{e}^{-t^2}dt$. $\int {\displaystyle \frac{2}{\sqrt{\pi }}}z{e}^{-z^2}dz$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:
$du={\displaystyle \frac{2}{\sqrt{\pi }}}{e}^{-z^2}$ 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:
$\int \text{erf }\left(z\right)dz=zerf\left(z\right)-\int {\displaystyle \frac{2}{\sqrt{\pi }}}z{e}^{-z^2}dz$
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$\int {\displaystyle \frac{2}{\sqrt{\pi }}}z{e}^{-z^2}dz$, for doing this let us consider $u=-z^2$ , so on differentiating it we get:
$\Rightarrow du=-2zdz$ and so
$\Rightarrow -\int {\displaystyle \frac{2}{\sqrt{\pi }}}z{e}^{-z^2}dz={\displaystyle \frac{1}{\sqrt{\pi }}}\int e^{u}du$
$\Rightarrow {\displaystyle \frac{1}{\sqrt{\pi }}}{e}^u+C$
$\Rightarrow {\displaystyle \frac{e^{-z^2}}{\sqrt{\pi }}}+C$
Putting it all together we get the final result which is:
$\int \text{erf }\left(z\right)dz=z\text{erf}\left(\text{z}\right)+{\displaystyle \frac{e^{-z^2}}{\sqrt{\pi }}}+C$

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.