Yes, that looks correct! Great job on finding a solution.

[Heimdall]

Hi,

I'm trying to calculate this integral :

$$\int_{-\infty}^{+\infty} e^{-\alpha x^2} tanh\left(x - \beta\right) dx$$

$$\alpha,\beta$$ being independant of x.

Would somebody know how to do that ?

Thanks :)

 

[jostpuur]

I don't know how to avoid the use of series, but this would be something with them:

Split the integral into two integrals, one over $$]-\infty,\beta]$$, and one over $$[\beta,\infty[$$. Then substitute the geometric series

$$\frac{1}{e^{x-\beta} + e^{-(x-\beta)}} = \frac{1}{e^{-(x-\beta)}}\sum_{n=0}^{\infty} (-1)^n e^{2n(x-\beta)},\quad\quad\quad x<\beta$$

and

$$\frac{1}{e^{x-\beta} + e^{-(x-\beta)}} = \frac{1}{e^{x-\beta}}\sum_{n=0}^{\infty} (-1)^n e^{-2n(x-\beta)},\quad\quad\quad x>\beta.$$

If I looked this right, now you should get such series for the integrand, that you know how to integrate each term in the series. Of course there's lot of work to be carried out, and in the end the result is in a form of series, so this is not the most desirable way to get the result... I'll be waiting eagerly to see if somebody has better ideas.

edit: oh no. I made one mistake. It is simple to integrate symmetric Gaussian peak over a domain $$]-\infty, 0]$$ or $$[0,\infty[$$, but actually splitting the integral of the original problem at $$\beta$$ results is something more difficult. I don't think my idea is working. But it could work if $$\beta=0$$.

edit edit: On the other hand, if $$\beta=0$$, then it is clear that the integral is zero, so actually I didn't help in anything

 

[Heimdall]

Hi,

thanks for trying :)

I think it would be helpful to rather consider this integral :

$$\int_{-\infty}^{+\infty} e^{-\alpha x^2}Erf\left(x - \beta\right)$$

Which is for the physicist that I am quite the same object..Edit :

Well in fact the real object I would like to evaluate is :

$$I\left(x\right) = \int_{-\infty}^{+\infty} e^{-\alpha v^2}Erf\left(v - A\left(x\right)\right) dv$$

Edit bis :

I could add that :

$$A\left(x\right) = Ln\left(cosh\left(x\right)\right)$$ But I think it doesn't matter what A(x) is to evaluate the integral..

 

[jostpuur]

Another strategy would be to try to use calculus of residues. It should be simple to integrate

$$\int\limits_{\gamma} e^{-\alpha z^2} \textrm{tanh}(z-\beta) dz$$

over any closed path, using the poles

$$\frac{1}{e^{z-\beta} + e^{-(z-\beta)}} \;=\; \frac{(-1)^n}{2}\frac{1}{z - (\beta + (\frac{1}{2} + n)\pi i)} \;+\; O\big(z - (\beta + (\frac{1}{2} + n)\pi i)\big), \quad\quad\quad n\in\mathbb{Z}$$

This alone doesn't yet solve the problem though, because one has to find out a way to deal with the arcs at infinity.

 

[Heimdall]

All right, here is a solution to my problem. As I said, to me $$Erf(x)$$ has the same behavior as $$tanh(x)$$, and it is indeed easier to integrate. Tell me if you agree :

$$I\left(A(x)\right) = \int_{-\infty}^{+\infty} e^{-\alpha v^2} Erf\left(v+A\left(x\right)\right)dv$$
is now the integral I'm looking to calculate.

First I can calculate the derivative acording to $$A(x)$$ :

$$I'\left(A(x)\right) =\frac{2}{\sqrt{\pi}} \int_{-\infty}^{+\infty} e^{-\alpha v^2}\frac{d}{dA}\left(\int_0^{v+A} e^{-t^2}dt\right) dv$$

$$I'\left(A(x)\right)=\frac{2}{\sqrt{\pi}} \int_{-\infty}^{+\infty} e^{-\alpha v^2}e^{-\left(v+A\right)} dv$$

this leads to :

$$I'\left(A(x)\right)= \frac{2e^{-\frac{\alpha}{\alpha+1}A(x)^2}}{\sqrt{\alpha+1}}$$Now integrating from 0 to A(x) gives :

$$I\left(A(x)\right) = \sqrt{\frac{\pi}{\alpha}}Erf\left(\frac{a}{a+1}A\left(x\right)\right)$$do you think that's correct ?