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

  • Thread starter ronphysics
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    Integral
In summary: You are trying to integrate x^2*exp(-x^2). You can try using the following substitution: y = x^2*exp(-x^2). Then use partial integration to get the infinite series.
  • #1
ronphysics
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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.
 
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  • #2
Did you get that right? [tex]x^2e^{x^2}[/tex] is divergent, those integrals will be [tex]\infty[/tex].
 
  • #3


alxm said:
Did you get that right? [tex]x^2e^{x^2}[/tex] is divergent, those integrals will be [tex]\infty[/tex].
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.
 
  • #4
Hmm, okay. Well what I'd do is first do a substitution, e.g.
[tex]y = x^2[/tex]
Resulting in:
[tex]\frac{1}{2}\int\sqrt ye^{-y}dy[/tex]
Then use partial integration which will land you with a infinite series for the indefinite integral.

For the case of:
[tex]\frac{1}{2}\int^\infty_0\sqrt ye^{-y}dy [/tex] you can 'cheat' and note that it's [tex]= \frac{1}{2}\Gamma(1.5) [/tex]
 
  • #5


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

For the case of:
[tex]\frac{1}{2}\int^\infty_0\sqrt ye^{-y}dy [/tex] you can 'cheat' and note that it's [tex]= \frac{1}{2}\Gamma(1.5) [/tex]
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.
 
  • #6
Hello ronphysics,

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

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

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

[tex]I=\int_{\beta}^{\infty}x^2e^{-x^2}dx[/tex]

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

[tex]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}[/tex]

Therefore, after rewriting:

[tex]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)[/tex]

Is this the method you were looking for?

coomast
 
  • #7
coomast said:
Hello ronphysics,

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

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

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

[tex]I=\int_{\beta}^{\infty}x^2e^{-x^2}dx[/tex]

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

[tex]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}[/tex]

Therefore, after rewriting:

[tex]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)[/tex]

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.
 
  • #8
Hi all,

I am trying for the following integration

∫x2 exp(a0+a1x+a2x2) dx

Please help me to find a solution.

Sivaji
 

1. What is the purpose of using erfc tables to evaluate the integral of x^2*exp(x^2)?

The complementary error function (erfc) tables provide a fast and accurate method for evaluating integrals involving the product of a polynomial and an exponential function. This method avoids the need for complex numerical integration techniques.

2. How do erfc tables work for evaluating the integral of x^2*exp(x^2)?

The erfc tables contain pre-calculated values of the complementary error function for a range of input values. By using interpolation techniques, the integral of x^2*exp(x^2) can be approximated by using these pre-calculated values.

3. Can erfc tables be used for other types of integrals?

Yes, erfc tables can be used for other types of integrals that involve the product of a polynomial and an exponential function. However, the accuracy of the approximation may vary depending on the specific integral being evaluated.

4. Are there any limitations to using erfc tables for evaluating integrals?

One limitation of using erfc tables is that they are only applicable for integrals involving the product of a polynomial and an exponential function. Other methods may be needed for evaluating integrals with different functions or more complex integrands.

5. How accurate are the results obtained from using erfc tables for evaluating the integral of x^2*exp(x^2)?

The accuracy of the results depends on the number of values in the erfc tables and the interpolation method used. Generally, the results obtained from using erfc tables are very accurate and can provide a good approximation for the integral of x^2*exp(x^2).

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