Question

How do you find the integral of ${e^{{x^2}}}$ ?

Answer 1

Hint: In the above question we have to find the integral of ${e^{{x^2}}}$ . As you know that the integration of an exponential function is the same. Moreover, in integration, we cannot integrate any constant number. So in the final answer, we add a constant. So let us see how we can solve this problem.

Step by step solution:
Let us find the integral of ${e^{{x^2}}}$ . This question cannot be solved or we can say that it has no finite solution. We will use an infinite series to solve this problem.
 $\Rightarrow {e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}}... = 1 + x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} + ...$ (for all x), it follows that,
 $\Rightarrow {e^{{x^2}}} = 1 + {x^2} + \dfrac{{{x^4}}}{2} + \dfrac{{{x^6}}}{6} + ...$ (for all x)
We will use the formula of $\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}}$ and we will integrate each of them.
 $\int {1 = x,} \int {{x^2}} = \dfrac{{{x^{2 + 1}}}}{{2 + 1}},\int {\dfrac{{{x^4}}}{2}} = \dfrac{{{x^{4 + 1}}}}{{2(4 + 1)}},\int {\dfrac{{{x^6}}}{6}} = \dfrac{{{x^{6 + 1}}}}{{6(6 + 1)}}$
Applying integration on both the sides
 $\Rightarrow \int {{e^{{x^2}}}} = \int {(1 + {x^2} + \dfrac{{{x^4}}}{2} + \dfrac{{{x^6}}}{6} + ...} )dx$
 $= C + x + \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{{10}} + \dfrac{{{x^7}}}{{42}} + ...$
We can see that we don’t get any feasible finite solution for the given problem.

Note:
In the above solution we solved the integration for ${e^{{x^2}}}$ but we get an infinite solution. According to Wolfram Alpha theorem, the antiderivative whose graph goes through the origin as $\dfrac{{\sqrt \pi }}{2}erfi(x)$ , where erfi(x) is called the "imaginary error function". Also, we have given the constant term in the integration as C.