Integral of x^2*exp(x^2) using erfc tables

[ronphysics]

Can anybody show me how to integrate x^2*exp(-x^2) between the limits 0 to infinity(symbol=00) and 1.5 to infinity with detail steps. I want this by using error function table. I know the 'multiplying integral method'. Here is what I did so far
int(0,00) x^2*exp(-x^2)
=x^2*int(0,00)exp(-x^2)-int(0,00)[2*x*int(0,00)exp(-x^2)]
=x^2(pi^0.5/2)-int(0,00)[2*x*(pi^0.5/2)] ...first term also between (0,00)
This seems to be giving 00! Where am I missing?
Thanks.

Also please show me how to do this between limits (1.5,00) with error function tables.

 

[alxm]

Did you get that right? $$x^2e^{x^2}$$ is divergent, those integrals will be $$\infty$$.

 

[ronphysics]

Did you get that right? $$x^2e^{x^2}$$ is divergent, those integrals will be $$\infty$$.

Oh, I am terribly sorry. I meant x^2*exp(-x^2). Forgot the negative sign. I corrected the original post (could not correct the Sub line). Thanks.

 

[alxm]

Hmm, okay. Well what I'd do is first do a substitution, e.g.
$$y = x^2$$
Resulting in:
$$\frac{1}{2}\int\sqrt ye^{-y}dy$$
Then use partial integration which will land you with a infinite series for the indefinite integral.

For the case of:
$$\frac{1}{2}\int^\infty_0\sqrt ye^{-y}dy$$ you can 'cheat' and note that it's $$= \frac{1}{2}\Gamma(1.5)$$

 

[ronphysics]

Hmm, okay. Well what I'd do is first do a substitution, e.g.
$$y = x^2$$
Resulting in:
$$\frac{1}{2}\int\sqrt ye^{-y}dy$$
Then use partial integration which will land you with a infinite series for the indefinite integral.

For the case of:
$$\frac{1}{2}\int^\infty_0\sqrt ye^{-y}dy$$ you can 'cheat' and note that it's $$= \frac{1}{2}\Gamma(1.5)$$

alxm, I don't know whether your approach is correct. But its pretty straightforward with 'multiplying integration'. You can find a method somewhere on the web.

My problem is trying to use error function approach.

 

[coomast]

Hello ronphysics,

You need to use the following four equations in order to solve it the way you want. These are:

$$\int_0^{\infty}e^{-ax^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}$$
$$erf(x)=\frac{2}{\sqrt{\pi}}\int_0^{x} e^{-t^2}dt$$
$$erfc(x)=\frac{2}{\sqrt{\pi}}\int_{x}^{\infty} e^{-t^2}dt$$
$$erf(x)=1-erfc(x)$$

Now I will show how to obtain the general question as:

$$I=\int_{\beta}^{\infty}x^2e^{-x^2}dx$$

Afterwards you need to find for yourself the solution to your integrals, and additionally check them by using the definite numbers for $\beta$. If any problems occur, just post. OK, so the first step is to use partial integration. This gives:

$$I=\int_{\beta}^{\infty}x^2e^{-x^2}dx= -\int_{\beta}^{\infty}\frac{x}{2}d\left(e^{-x^2}\right)= -\left[\frac{x}{2}e^{-x^2}\right]_{\beta}^{\infty}+\frac{1}{2}\int_{\beta}^{\infty}e^{-x^2}dx= \frac{\beta}{2}e^{-{\beta}^2}+\frac{1}{2}erfc(\beta)\frac{\sqrt{\pi}}{2}$$

Therefore, after rewriting:

$$I=\frac{\beta}{2}e^{-{\beta}^2}+\frac{\sqrt{\pi}}{4}erfc({\beta})= \frac{\beta}{2}e^{-{\beta}^2}+\frac{\sqrt{\pi}}{4}\left(1-erf({\beta})\right)$$

Is this the method you were looking for?

coomast

 

[ronphysics]

Hello ronphysics,

You need to use the following four equations in order to solve it the way you want. These are:

$$\int_0^{\infty}e^{-ax^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}$$
$$erf(x)=\frac{2}{\sqrt{\pi}}\int_0^{x} e^{-t^2}dt$$
$$erfc(x)=\frac{2}{\sqrt{\pi}}\int_{x}^{\infty} e^{-t^2}dt$$
$$erf(x)=1-erfc(x)$$

Now I will show how to obtain the general question as:

$$I=\int_{\beta}^{\infty}x^2e^{-x^2}dx$$

Afterwards you need to find for yourself the solution to your integrals, and additionally check them by using the definite numbers for $\beta$. If any problems occur, just post. OK, so the first step is to use partial integration. This gives:

$$I=\int_{\beta}^{\infty}x^2e^{-x^2}dx= -\int_{\beta}^{\infty}\frac{x}{2}d\left(e^{-x^2}\right)= -\left[\frac{x}{2}e^{-x^2}\right]_{\beta}^{\infty}+\frac{1}{2}\int_{\beta}^{\infty}e^{-x^2}dx= \frac{\beta}{2}e^{-{\beta}^2}+\frac{1}{2}erfc(\beta)\frac{\sqrt{\pi}}{2}$$

Therefore, after rewriting:

$$I=\frac{\beta}{2}e^{-{\beta}^2}+\frac{\sqrt{\pi}}{4}erfc({\beta})= \frac{\beta}{2}e^{-{\beta}^2}+\frac{\sqrt{\pi}}{4}\left(1-erf({\beta})\right)$$

Is this the method you were looking for?

coomast

coomast,

Thanks. Thats exactly what I was looking for. My mind was stuck in separating variables as x^2 and exp(-x^2). However, the useful way to seprate is x and x*exp(-x^2). Thanks again.

 

[sivaji]

Hi all,

I am trying for the following integration

∫x² exp(a0+a1x+a2x²) dx

Please help me to find a solution.

Sivaji