# Integral of squared error function

1. Apr 9, 2010

### singhofmpl

Does anybody knows how to solve the following equation?
$$\int_{0}^{\infty}\mbox{erf}^2(\sqrt{x})\exp(-x)dx$$

regards
singhofmpl

2. Apr 9, 2010

### Gib Z

Policy is that you must show us working before we can provide substantial help. If we don't know what you have done we done know where you got stuck. The hints I'll give you this time are: Trying integrating by parts to reduce the exponent of the error function in the integral. You can find the derivative of the error function by the Fundamental Theorem of Calculus.

3. Apr 9, 2010

### singhofmpl

Dear Sir
Your suggestion is very much correct as far as applying by parts is concerned but I'm able to get analytical results only if the integral involves only one power of the error function i.e

$$\int_{0}^{\infty}\mbox{erf}(\sqrt{x})\exp(-x)dx$$ but I'm not able to get the correcr results for the integral $$\int_{0}^{\infty}\mbox{erf}^2(\sqrt{x})\exp(-x)dx$$. If I apply by parts to the integral involving squared error function then after a long derivation finally I'm getting 0, which I' m suspicious that this solution is correct.

4. Apr 9, 2010

### Gib Z

The answer is not zero as the integrand is strictly positive on the interval of integration. Please show me the steps you did. From the start: To reduce the power of $$Erf^2\sqrt{x}$$ in your integral, use integration by parts, letting $$u= \Erf^2 \sqrt{x}, dv = e^{-x} dx$$. To find du, use the definition of the Error function and the Fundamental Theorem of Calculus.

After you carry that out, you should have an integral like : $$\int^{\infty}_0 Erf (\sqrt{x}) e^{-2x} dx$$

So exp (-2x), not just exp (-x).

You can't finish there. You have to apply integration by parts again in a similar fashion, then continue from there.

Last edited: Apr 9, 2010
5. Apr 9, 2010

### singhofmpl

Respected Sir
I'm sorry to say bet still there is problem. Please see my workings given below:
$$\int_{0}^{\infty}\mbox{erf}^2(\sqrt{x})\exp(-x)dx$$
let $$u=\mbox{erf}^2(\sqrt{x})$$
Now $$du=\frac{2\mbox{erf}(\sqrt{x})\exp(-x)}{\sqrt{\pi}\sqrt{x}}dx$$

Now it is required to write $$\exp(-x)dx$$ in terms of $$du$$. So
$$\exp(-x)dx=\frac{\sqrt{\pi}\sqrt{x}}{2\mbox{erf}(\sqrt{x})}du$$
which further can be written as:
$$\exp(-x)dx=\frac{\sqrt{\pi}\sqrt{x}}{2\sqrt{u}}du$$
but the problem is how to write $$\sqrt{x}$$ in terms of u?

6. Apr 9, 2010

7. Apr 9, 2010

### singhofmpl

Dear Sir
Thanks a lot for your prompt response. I'm really very sorry for bothering you so much. Finally I got the solution following your advice. The final answer which I'm getting is given below:

$$\frac{2\sqrt{2}\arctan(\frac{1}{\sqrt{2}})}{\pi}$$

8. Apr 9, 2010

### Gib Z

That is correct. Don't worry about asking questions, It wasn't a bother at all.

Last edited: Apr 9, 2010
9. Apr 9, 2010

### singhofmpl

Dear Sir
May I know your affiliation (if you don't mind) so that I can mention it while writing my report or paper?

10. Apr 9, 2010

### Gib Z

My name is Ragib Zaman, and I am a first year student at the University of Sydney, Australia. I am studying the Bachelor of Science (Advanced Mathematics) degree.

11. Jun 6, 2010

### singhofmpl

Integral due to non-linearity of PA

Hi Ragib
While modeling the non-linear behavior of power amplifier (PA), I came to an integral of the form given below; I'm not able to figure out how to solve it.

$$I=\frac{A^2}{2\sigma_x^4}\int_{0}^{\infty}\frac{r^3}{r^2+A^2}\exp(\frac{j\pi}{3}\frac{r^2}{r^2+A^2}-\frac{r^2}{2\sigma_x^2})dr$$

12. Jun 6, 2010

### Gib Z

After some initial manipulations I've managed to make it look like

$$ke^k \int^{\infty}_k \left(1-\frac{k}{t}\right) \frac{e^{\frac{i\pi}{3}\left(1-\frac{k}{t}\right)}}{e^t} dt$$

where $$k = \frac{A^2}{2\sigma_x^2}$$

I'm still working on reducing that further, but in the mean time you could give that form a try.

13. Jul 9, 2010

### bapi

Dear Sir
kindly suggest me how find the integration of :

$$\int_{0}^{\infty}\mbox{erf}(\sqrt{x})\exp(-x)dx$$

14. Oct 31, 2010

### meng

Hi, may I know how do you solve for $$\int_{0}^{\infty}\mbox{erf}(\sqrt{x})\exp(-x)dx$$?

Could you detail the assignment of 'u' in each iteration of integration, and how do you get the eventual term in arctan?

Could you give me some hints on solving that because it seems to me that the integrations would loop infinitely.

15. Oct 31, 2010

### Gib Z

It's not very hard, I'm certain you can do it if you try. Integration by parts with $u = erf( \sqrt{x} ), dv = e^{-x} dx$. To find du, use the fundamental theorem of calculus and the chain rule.

Last edited: Oct 31, 2010
16. Nov 1, 2010

### meng

Regarding the original problem:
$$\int_{0}^{\infty}\mbox{erf}^2(\sqrt{x})\exp(-x)dx$$

here is what I've got so far.
$$u=\mbox{erf}^2(\sqrt{x}),$$
$$du=\frac{2\mbox{erf}(\sqrt{x})\exp(-x)}{\sqrt{\pi}\sqrt{x}}dx,$$
$$dv = \exp(-x) dx,$$
$$v=-\exp(-x).$$

Thus, we have
$$\int_{0}^{\infty}\mbox{erf}^2(\sqrt{x})\exp(-x)dx = -\exp(-x)\mbox{erf}^2(\sqrt{x}) + \int_{0}^{\infty}\frac{2\mbox{erf}(\sqrt{x})\exp(-2x)}{\sqrt{\pi}\sqrt{x}}dx.$$

Now suppose that I do another integration by parts with
$$u=\mbox{erf}(\sqrt{x}),$$
$$du=\frac{\exp(-x)}{\sqrt{\pi}\sqrt{x}}dx,$$
$$dv = \frac{\exp(-2x)}{\sqrt{x}} dx,$$
$$v=\frac{\sqrt{\pi}\mbox{erf}(\sqrt{2x})}{\sqrt{2}}.$$

The resultant expression becomes
$$\int_{0}^{\infty}\mbox{erf}^2(\sqrt{x})\exp(-x)dx = -\exp(-x)\mbox{erf}^2(\sqrt{x}) + \sqrt{2}\mbox{erf}(\sqrt{x})\mbox{erf}(\sqrt{2x}) - \frac{2\sqrt{2}}{\sqrt{\pi}}\int_{0}^{\infty}\frac{\mbox{erf}(\sqrt{2x})\exp(-x)}{\sqrt{x}}dx.$$

and I'm stuck at this point where the form of $$\frac{\exp(x)\mbox{erf}(x)}{x}$$ in the integration cannot be suppressed.

I have tried to proceed with another round of integration and a similar form returns. Please kindly point me out my mistake(s). Thanks.

Last edited: Nov 1, 2010
17. Nov 17, 2010

### Gib Z

Sorry about the very late reply meng, I don't come on physicsforums very often anymore because of my studies. To others reading, I'm reviving this thread because I got a PM which stated this problem was in their research, so I thought I should help.

Here you go meng:

$$I = \int^{\infty}_0 \mbox{erf}(\sqrt{x}) e^{-x} dx$$

Integrate by parts, letting $u = \mbox{erf} \sqrt{x} , du = \frac{1}{2\sqrt{x}} \cdot \frac{2}{\sqrt{\pi}} e^{-x} dx$ and $dv = e^{-x} dx, v = - e^{-x}$, where we found the derivative of u by the fundamental theorem of calculus.

This gives:
$$I = \left[ -\mbox{erf}\sqrt{x} e^{-x} \right]^{\infty}_0 + \int^{\infty}_0 \frac{ e^{-2x} }{ \sqrt{\pi x} } dx$$.

For the 1st term, as x goes to infinity, the erf term goes to a constant since $\int^{\infty}_0 e^{-x^2} dx = \sqrt{\pi}/2$ while the exp goes to zero, so thats zero. And when x=0, the erf term is 0 while the exp term is 1, so zero as well. So we just have the integral remaining, in which we let $u = \sqrt{x}$ which leaves us with

$$I = \frac{2}{\sqrt{\pi}} \int^{\infty}_0 e^{-2u^2} du$$.

Let $t = \sqrt{2} u$,

$$I = \frac{ \sqrt{2}}{\sqrt{\pi}} \int^{\infty}_0 e^{-t^2} dt = \frac{1}{\sqrt{2}}$$

18. Jul 19, 2011

### xrod

Here's a very late reply. I find it a little easier to change variables so that the error function doesn't have the square root:
$$y^2 = x$$
$$I = \int_{0}^{\infty}\mbox{erf}^2(\sqrt{x})\exp(-x)dx = 2 \int_0^\infty dy~\mbox{erf}^2(y) \exp(-y^2) y dy$$
Now integrate by parts like you did, the surface term is zero
$$I = \frac{4}{\sqrt{\pi}} \int_0^\infty dy ~\mbox{erf}(y) \exp(-2y^2)$$
At this point substitute the definition of the error function but with a change of variables so that the integration limits are constant
$$\mbox{erf}(y) = \frac{2}{\sqrt{\pi}}\int_0^1 d\alpha ~e^{-\alpha^2 y^2} y$$
You can then change the order of integration
$$I = \frac{8}{\pi} \int_0^1 d\alpha \int_0^\infty dy ~ y \exp[-(2+\alpha^2)y^2] = \frac{4}{\pi} \int_0^1 d\alpha ~\frac{1}{2+\alpha^2} =\frac{2\sqrt{2}}{\pi} \tan^{-1}\frac{1}{\sqrt{2}}$$