Approximating step function with Tanh to compute integral

Question

It is easy to compute by hand the following integral.

$$ Q(k) = \lim \limits_{\varepsilon \to 0} \int_2^{\infty} \frac{e^{-\varepsilon r}}{k}\sin(kr) \mathrm{d}r = \frac{\cos(2k)}{k^2}$$

I wanted to use smoothened version of step function to compute this integral using Mathematica. My idea was to rewrite this integral in the following way.

$$ Q(k) = \lim \limits_{\varepsilon \to 0} \int_0^{\infty} \Theta(r-2) \frac{e^{-\varepsilon r}}{k}\sin(kr) \mathrm{d}r$$

Here, I use step function $\Theta(r-2)$ in the following smooth form.

$$ \Theta(r-2) = \frac{1}{2} + \frac{1}{2} \tanh(1000(r-2)) $$

I would like to compute this integral in this way and then compare it to the result $\cos(2k)/k^2$. To do the computations, I use the following code.

Regularized[k_, epsilon_] := Assuming[epsilon >= 0 && k >= 0, Integrate[Sin[x*k]/k*Exp[-epsilon*x]*(0.5 + 0.5*Tanh[1000*(x - 2)]), {x, 0, Infinity}]]; 

ft[k_] := Limit[Regularized[k, P], P -> 0, Direction -> "FromAbove"];

To compare, I now do two plots using the following commands.

Plot[ft[k], {k, 0, 10}]

Plot[Cos[2 k]/(k*k), {k, 0, 10}]

My problem is that this computation takes more than 2 hours on my PC (which is not that old), and I have not yet seen the final result, so I wonder if I do something wrong. Maybe this approximation for step function is bad and instead of $\tanh$ I should use something else? Or is there a problem with my Mathematica code? Any help is very much appreaciated

Answer 1

Side-stepping the main question some, Rubi is capable of integrating this much more directly. In my own experience, adding a smoothed function to simplify a symbolic problem requires ensuring that the function in question has convenient cancellation behavior. I can't say that the choice of $\tanh$ is good or bad in this case, but it certainly does take longer than I'd expect on Mathematica.

On the other hand, using Rubi:

<< Rubi`
expr = Int[Exp[-ε r]/k Sin[k r], {r, 2, Infinity}]

ConditionalExpression[(E^(-2 ε) Cos[2 k])/(k^2 + ε^2) + (E^(-2 ε) ε Sin[2 k])/(k (k^2 + ε^2)), k > 0 && ε > 0]

If you plug this expression into an appropriate Limit:

Limit[expr, ε -> 0, Direction -> -1, Assumptions -> {k > 0}]

ConditionalExpression[Cos[2 k]/k^2, k > 0]

It also turns out that the original integrand also works if k is explicitly assumed to be positive:

 expr2 = Integrate[Exp[-ε r]/k Sin[k r], {r, 2, Infinity}, Assumptions -> {ε > 0, k > 0}]

Using HeavisideTheta also works as a step function under the same assumptions:

 Integrate[HeavisideTheta[r - 2] Exp[-ε r]/k Sin[k r], {r, 0, Infinity}, Assumptions -> {ε > 0, k > 0}]

Both of these expressions can be subjected to the same limit as above to give the same result. Again, I would expect that introducing inexact numbers or additional complicated functions would tend to make the result take longer, unless the form is peculiarly chosen to simplify the problem tremendously. I also generally trust that if a closed form solution to an integral exists and is at least somewhat well known, Rubi can provide a good head start on figuring out what it is.

Answer 2

There is another approximation for the step function which is commonly used in AI, namely the logistic function which takes the generic form $1/(1+e^{-x})$. In particular, it has the property $$\vphantom{``}%The editor complained at me unless I put this invisible code here... \Theta(x)=\lim_{z\rightarrow\infty}\frac{1}{1+e^{-zx}}. $$ For the particular case of $\Theta(x-2)$ that you're interested in, we can compare the approximation you had using the $\tanh$ to the logistic function with $z=50$ (chosen such that the approximation isn't so good that we can't tell the functions apart):

Plot[{(0.5+0.5*Tanh[1000*(x-2)]),1/(1+Exp[-50*(x-2)])},{x,1,3}]

Now, the advantage of the logistic map in AI is typically that it's very fast to evaluate, so there are numerical advantages. It seem like here you're looking for a symbolic result (as you're using Integrate rather than NIntegrate, which admittedly fails to converge for the $\tanh$ approximation), so numerical efficiency isn't so helpful, but because this approximation seems to give a nicer result. In fact, the integral $$ f(k;\epsilon,z)=\int_0^\infty\frac{\sin(kx)}{k}e^{-\epsilon x}\frac{1}{1+e^{-z(x-2)}} $$ turns out to be doable in closed form for all $\epsilon,k,z>0$. Adding the $z>0$ condition to the assumptions in your original integration, Mathematica evaluates this integral to $$ f(k;\epsilon,z)=\frac{1}{2k(\epsilon^2+k^2)}\left[(k+i\epsilon)_2F_1\left(1,\frac{\epsilon+ik}{z};\frac{\epsilon+ik+z}{z};-e^{2z}\right)+(k-i\epsilon)_2F_1\left(1,\frac{\epsilon-ik}{z};\frac{\epsilon-ik+z}{z};-e^{2z}\right)\right], $$ where $_2F_1$ is one of the hypergeometric functions. Plotting this function for $\epsilon=0.05$ and $z=100$, we can see

Plot[{f[k,epsilon,z] /. {epsilon -> 0.05, z -> 100}, Cos[2*k]/k^2}, {k, 0, 10},PlotRange -> {{0, 10}, {-0.2, 0.2}}, PlotLegends -> Automatic]

where the yellow curve is the exact result and the blue curve is our approximation.