What is $\int_{0}^{\infty} e ^{x^2} dx$

Question

I am working in some economic problems and got in stuck with the following integral

What is $\int_{0}^{\infty} e ^{x^2} dx$?

Actully, I remember that we should use Taylor series to calculate it but I do not know more precisely.

Could you please show me in detail?

Thank you in advance.

Answer 1

\begin{align} & \int_0^\infty e^{x^2} \, dx \\[10pt] \ge {} & \int_0^\infty 1 \, dx \text{ because } e^{x^2}>1 \text{ when } x^2>0 \\[10pt] = {} & +\infty \end{align}

Answer 2

Hint: Consider $I := \int_{-\infty}^{\infty} e^{-x^2}dx$. Then notice $I^2 =\int_{-\infty}^{\infty} e^{-x^2}dx\int_{-\infty}^{\infty} e^{-y^2}dy $.

Now convert the above expression into polar coordinates and solve for $I^2$. Then take the square root of $I$, then divide $I$ by 2 to obtain the desired solution (notice we can divide by 2 since our $e^{-x^2}$ is even, so $ \int_{-\infty}^{\infty} e^{-x^2}dx = 2 \int_{0}^{\infty} e^{-x^2}dx$

Edit: for the explicit details: $\int_{-\infty}^{\infty} e^{-x^2}dx\int_{-\infty}^{\infty} e^{-y^2}dy $= $\int_{-\infty}^{\infty} dx\int_{-\infty}^{\infty} e^{-(y^2+x^2)}dy $. = $\int_{\theta=0}^{2 \pi} d\theta\int_{r=0}^{\infty} re^{-r^2}dr$

Incase you mean $\int_{0}^{\infty} e^{x^2}$, it is defined to be $\lim_{a \rightarrow \infty} \int_0^a e^{x^2} dx$. In particular, this integral is $\infty$. To see this notice that $e^{x^2} - x^2 \geq 0$ for every $x \in \mathbb{R}$. Hence, we must have $\int_0^a e^{x^2} dx \geq \int_{0}^{a} x^2 dx$. Then as $a \rightarrow \infty$, the integral of $x^2$ goes to infinity, hence this means $\lim_{a \rightarrow \infty} \int_0^a e^{x^2} dx = \infty$ as desired.

Answer 3

$$\int_0^\infty e^{x^2}\,dx > \int_0^\infty x^2 \,dx = \left. \frac{1}{3}x^3 \right|_0^\infty = \infty$$

Unless you meant $\int_0^\infty e^{-x^2}\,dx$ but this is a rather famous integral!

Answer 4

Considering that $$I_p=\int_{0}^{p} e ^{x^2} dx=e^{p^2} F(p)$$ where appears the Dawson integral. If you look here, you will see that, for large values of $p$, $$F(p)= \sum_{k=0}^{\infty} \frac{(2k-1)!!}{2^{k+1} p^{2k+1}} = \frac{1}{2 p}+\frac{1}{4 p^3}+\frac{3}{8 p^5}+O\left(\frac{1}{p^7}\right)$$ This gives a quite good approximation of $I_p$ as shown below $$\left( \begin{array}{ccc} p & \text{approximation} & \text{exact} \\ 5 & 7.353140320\times 10^9 & 7.354153748\times 10^9 \\ 6 & 3.644670574\times 10^{14} & 3.644831078\times 10^{14} \\ 7 & 1.376717922\times 10^{20} & 1.376741346\times 10^{20} \\ 8 & 3.928126796\times 10^{26} & 3.928156311\times 10^{26} \\ 9 & 8.419813202\times 10^{33} & 8.419844074\times 10^{33} \\ 10 & 1.350879668\times 10^{42} & 1.350882281\times 10^{42} \\ 20 & 1.307005251\times 10^{172} & 1.307005289\times 10^{172} \\ 30 & 1.222148762\times 10^{389} & 1.222148765\times 10^{389} \\ 40 & 9.294304168\times 10^{692} & 9.294304173\times 10^{692} \\ 50 & 5.448684489\times 10^{1083} & 5.448684490\times 10^{1083} \end{array} \right)$$ Using the above and series, you could show that $$\log(I_p)=p ^2+\log\left(\frac 1{2p} \right)+\frac{1}{2 p^2}+\frac{5}{8 p^4}+\frac{37}{24 p^6}+O\left(\frac{1}{p^8}\right)$$