I have to integrate $$\int e^{x^2}dx$$ I tried using product rule and saw that this will form an infinite series.
Now I have two questions
- How to integrate this function?
2.If this is not integrable, then I have some more questions which are behaving same or not being solved by me . How can I know those function are integrable or not?
On
The function $e^{x^2}$ is continuous on $\mathbb{R}$ so it is integrable in the sense that the definite integral $$\int_a^b e^{x^2} \,\mbox{d}x$$ exists for all real numbers $a,b$. The fundamental theorem of calculus also guarantees the existence of a primitive function (anti-derivative) of $e^{x^2}$.
But what you may be looking for is a closed form, elementary anti-derivative. That does not exist for $e^{x^2}$ so you'll either need to work with series or 'hide' the problem by using special functions such as the (Gauss) error function.
(...) but does there exists any function which are not integrable??
If you're looking for a function that isn't (Riemann) integrable, then this is a classic example: $$f(x)=\begin{cases} 1 &\text{if } x\in \mathbb Q \\ 0 & \text{if } x\notin \mathbb Q \end{cases}$$
On
In terms of known functions a primitive for your integral can be written in terms of the so-called imaginary error function $\text{erfi} (x)$. Defined as $$\text{erfi} (x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{t^2} \, dt,$$ we see that $$\int e^{x^2} \, dx = \frac{\sqrt{\pi}}{2} \text{erfi} (x) + C.$$
Note that despite the name imaginary error function, $\text{erfi} (x)$ is real when $x$ is real.
It's integrable but the only way you can express this easily is $$ F\left(x\right)=\sum_{n=0}^{+\infty}\frac{x^{2n+1}}{n!\left(2n+1\right)} $$