How to find $\int \frac{e^{-x^2}}{x^2 + 1} dx$?

Question

I have a question about improper integrals:

How can we find $\lim_{n \rightarrow +\infty}\int_{-n}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx$?

$\textbf{Some effort:}$

$\lim_{n \rightarrow +\infty}\int_{-n}^{n} \frac{1}{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx = \lim_{n \rightarrow +\infty} \frac{2}{n}\int_{0}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx $ $~~~~~~~~~\textbf{(1)}$

By setting $nx^2 = u$, we have $dx = \frac{1}{2\sqrt{n}} \times \frac{1}{\sqrt{u}}$ and $x = \frac{\sqrt{u}}{\sqrt{n}}$. So by substituting these in $\textbf{(1)}$ we have (I will not put bounds and at the end will come back to the initial bounds and also I will drop the constant in integrals)

$\textbf{(1)} = \lim_{n \rightarrow +\infty} \frac{2}{n} \int \frac{1 - e^{-u}}{\frac{u}{n}(1+u)} \times \frac{1}{2 \sqrt{n}}\times \frac{1}{ \sqrt{u}} du = \lim_{n \rightarrow +\infty} \frac{1}{\sqrt{n}} \int \frac{1 - e^{-u}}{ u \sqrt{u}(1+u)} du $

$= \lim_{n \rightarrow +\infty} \frac{1}{\sqrt{n}} \int ( \frac{1 }{ u \sqrt{u}(1+u)} - \frac{e^{-u} }{ u \sqrt{u}(1+u)}) du$ $~~~~~~~~~\textbf{(2)}$

By setting $\sqrt{u} = v$, we have $\frac{1}{2\sqrt{u}}du = dv$. So by substituting these in $\textbf{(2)}$ we have

$\textbf{(2)} = \lim_{n \rightarrow +\infty} \frac{2}{\sqrt{n}} \int (\frac{1}{v^2(1+v^2)} - \frac{e^{-v^2}}{v^2(1+v^2)} dv) $ $~~~~~~~~~\textbf{(3)}$

$\textbf{(3)}= \lim_{n \rightarrow +\infty} \frac{2}{\sqrt{n}} \int ( \frac{1}{v^2} - \frac{1}{1+v^2} - \frac{e^{-v^2}}{v^2} + \frac{e^{-v^2}}{v^2 + 1}) dv$

Now we will calculate each term separately.

First part:

For to find $\int \frac{1}{v^2} dv $, by setting $k_1=-\frac{1}{v}$, we have $dk_1 =\frac{1}{v^2} dv$ and so we have $\int \frac{1}{v^2} dv = \int dk_1= k_1= -\frac{1}{v}= -\frac{1}{\sqrt{u}}= -\frac{1}{\sqrt{nx^2}} = -\frac{1}{\sqrt{n}|x|} $

Second part:

For to find $-\int \frac{1}{1+v^2} dv$, by setting $k_2 = \arctan(v)$, we have $dk_2 = \frac{1}{1 + v^2}dv$ and so we have $-\int \frac{1}{1+v^2} dv = -int dk_2= -k_2= -\arctan(v) = -\arctan(\sqrt{u}) = -\arctan(\sqrt{nx^2}) = -\arctan(\sqrt{n}|x|) $

Third part:

For to find $-\int \frac{e^{-v^2}}{v^2} dv $, by setting $\begin{cases} k_3=e^{-v^2}\\ -\frac{1}{v^2}=dk_4 \end{cases}$ we will have $\begin{cases} dk_3=-2ve^{-v^2}\\ k_4= \frac{1}{v} \end{cases}$ and our integral will transform to $-\int \frac{e^{-v^2}}{v^2} dv = \frac{e^{-v^2}}{v} - 2 \int e^{-v^2} dv = \frac{e^{-v^2}}{v} - 2(\frac{\sqrt{\pi}}{2}) = \frac{e^{-v^2}}{v} - \sqrt{\pi} = \frac{e^{-(\sqrt{u})^2}}{\sqrt{u}} - \sqrt{\pi} = \frac{e^{-u}}{\sqrt{u}} - \sqrt{\pi} =\frac{e^{-nx^2}}{\sqrt{nx^2}} - \sqrt{\pi}=\frac{e^{-nx^2}}{\sqrt{n}|x|} - \sqrt{\pi}$

Forth part:

For to find $\int \frac{e^{-v^2}}{v^2 + 1} dv$, I cannot find it!

Can someone please help me to find $\int \frac{e^{-v^2}}{v^2 + 1} dv$?

Thanks!

Answer 1

$$I_n=\int_{-n}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}\,dx=2\int_{0}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}\,dx$$ Let $$nx^2=t\implies x=\frac{\sqrt{t}}{\sqrt{n}}\implies dx=\frac{dt}{2 \sqrt{n} \sqrt{t}}$$ making $$I_n=\sqrt{n}\int_0^1\frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}\,dt$$ You do not need to compute anything else to show that $I_n$ is just proportional to $\sqrt{n}$ and just conclude.

Now, as said in answers, the last integral cannot be computed and series expansions are required. Using Taylor, we would have $$\frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}=\frac{1}{\sqrt{t}}-\frac{3 \sqrt{t}}{2}+\frac{5 t^{3/2}}{3}-\frac{41 t^{5/2}}{24}+\frac{103 t^{7/2}}{60}-\frac{1237 t^{9/2}}{720}+O\left(t^{11/2}\right)$$ $$\int \frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}\,dt=2 \sqrt{t}-t^{3/2}+\frac{2 t^{5/2}}{3}-\frac{41 t^{7/2}}{84}+\frac{103 t^{9/2}}{270}-\frac{1237 t^{11/2}}{3960}+O\left(t^{13/2}\right)$$ Using the bounds, we should get $$\int \frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}\,dt=\frac{1634621}{1081080}\approx 1.51203$$ while numerical integration would lead to $\approx 1.38990$

Answer 2

Unfortunately, this integral has no closed form in terms of known functions, elementary or non-elementary. You'd have to integrate it numerically if you want to do anything practical with it.

The series expansion of the integral (at $x=0$), while by no means a closed form, is, according to W|A:

$$\int \frac{e^{-v^2}}{v^2 + 1} dv \approx x - \frac{2 x^3}3 + \frac{x^5}2 - \frac{8x^7}{21} + \frac{65 x^9}{216} - \frac{163 x^11}{660} + O(x^{13}) $$

And of your original integral:$$\int \frac{e^{-nx^2}}{x^2(1 + nx^2)} dx \approx -\frac 1x - 2 n x + \frac{5 n^2 x^3}6 - \frac{8 n^3 x^5}{15} + O(x^6) $$