5

In a now deleted post appeared an interesting integral.

$$I=\int\frac 1 x \,\,\sin \left(\frac{\log (x)}{x}\right) \,dx$$ which does not make (too much) problems from a numerical point of view.

As one can expect, the plot of the integrand is not the most pleasnat we could find.

$$\frac{\log(x)}x=-t \quad\implies\quad x=\frac{W(t)}{t}\quad \implies \quad $$ $$I=\int \frac{W(t) }{1+W(t)}\,\, \frac {\sin(t)} t \, dt=\text{Si}(t)-\int \frac{1 }{1+W(t)}\,\, \frac {\sin(t)} t \, dt$$ which is much more pleasant to look at and even easier to integrate nmerically.

My question is : how could we approximate the integrand to have a decent approximation for the integral between $k\pi$ and $(k+1)\pi$ ?$k$ being a non negative integer.

Edit

Thanks to @Hume2's answer, the problem now reduces to $$I=W(t)\,\sin(t)-\int W(t)\,\cos(t)\, dt$$ $$J_k=-\int_{k\pi}^{(k+1)\pi} W(t)\,\cos(t)\, dt$$

Asymptotically, it seems that $$|J_k| \sim \frac{1}{2 k}+\frac{3}{4k^2}+\frac{19}{4k^3}+O\left(\frac{1}{k^4}\right)$$

Claude Leibovici
  • 289,558
  • Is $k>0$ ? Is it an integer or any real number ? – Sam May 01 '24 at 05:27
  • 1
    @Sam. If you plot the function, I suppose that you will see that the interesting part is between the zero's of the integrand. So, yes, $k$ has to be a non negative integer. Thanks for pointing. Cheers :-) – Claude Leibovici May 01 '24 at 05:32
  • Just spotted your question, Claude. It happens that I have used the relationship given by @Hume2 (under a slightly different form) in a recent answer of mine here to an interesting question : see equation (7) of my "second method"... – Jean Marie May 04 '24 at 14:50

1 Answers1

5

According to Wolfram Alpha:$$ \int\frac{W(t)}{(1+W(t))t}=W(t)+C $$ So we can apply per-partes to the integral and we get:$$ I=W(t)\sin(t)-\int W(t)\cos(t)\mathrm dt $$ If you are interested only in integrals between multiples of $\pi$, then the term before the integral cancels out. Can you continue from now on?

Hume2
  • 2,759