What is the integral of the error function?
Answer
597.6k+ views
Hint: To solve this question we need to know the concept of integration. The error function is denoted by erf, which defined as $\text{erf z = }\dfrac{2}{\sqrt{\pi }}\int\limits_{0}^{z}{{{e}^{-{{t}^{2}}}}dt}$. 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.
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.
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