# When does the sine transform result in a positive function?

For each $$x>0$$ is, $$\int_0^\infty \tanh(t)e^{-\cosh(t)} \sin(x t) dt > 0\ \ \ ?$$

In general, it is known that if the kernel is decreasing, then the sine transform is positive. Note that this is not a decreasing kernel, hence for this example, this theorem would not apply. Any references or proofs would be helpful.

UPDATE

Even though this problem is entirely composed of somewhat elementary functions, it appears this problem is certainly non-trivial. It appears a good reference for Fourier inequalities should probably be created/compiled at some point. In any case, I have attempted two possible approaches for this particular example, but am unsure if they are useful, and am unsure where to proceed.

METHOD 1:

I have noticed that one can split $$\tanh(t) = \coth(2t)-\text{csch}(2t)$$ creating a singularity at zero. I am sure countless other splits also exist. This means that $$\int_0^\infty \tanh(t) e^{-\cosh(t)} \sin(xt) =$$ $$\int_0^\infty \coth(2t) e^{-\cosh(t)} \sin(xt) - \int_0^\infty \text{csch}(2t) e^{-\cosh(t)} \sin(xt)$$

Both of these kernels are now decreasing. This means we have a difference of positive functions. One possible thought is that the derivative of the left is always larger than the derivative of the right, making the left always larger. Haven't explored this yet (if it is even true). This approach may appear to be unhelpful.

METHOD 2:

Another thought is that the dirichlet series of $$\tanh$$, valid for $$t>0$$, is $$\tanh(t) = \sum_{k=0}^\infty (-1)^k (1-e^{-2t})e^{-2kt}$$ Hence it might be helpful to analyze $$\sum_{k=0}^\infty (-1)^k \int_0^\infty (1-e^{-2t})e^{-2kt-\cosh(t)} \sin(xt)$$ Unfortunately, these are not decreasing kernels, but do appear to also have positive sine transforms (why?). Moreover, they are not decreasing with respect to each other either (they don't form a alternating decreasing sequence), for example, $$x=10$$,

k=0   0.00149873
k=1  -0.00382987
k=2   0.00479947
k=3  -0.00468230
k=4   0.00405178
k=5  -0.00330839
k=6   0.00263354


The subtlety of these kernels is very difficult to understand. Take for example, $$\int_0^\infty (K_1:=(1-e^{-2t})e^{-\cosh(t)}) \sin(10t) = 0.00149873$$ $$\int_0^\infty (K_2:=(1-e^{-2t})^2 e^{-\cosh(t)}) \sin(10t) = -0.00233114$$ $$\int_0^\infty (K_3:=(1-e^{-2t}) e^{-2t-\cosh(t)}) \sin(10t) = 0.00382987$$ Multiplying by $$e^{-2t}$$ appears to increase the transform; whereas, $$1-e^{-2t}$$ appears to decreasing the transform, even causing it to be negative. Both factors are less than 1, but the later is increasing, where the former is decreasing. Is there a generality to know here?

• The function $f(t) = \tanh(t)e^{-\cosh(t)}$ is the derivative of $-g(t)$, where $g(t) = Ei(-\cosh(t))$. Thus, the sine transform of $f$ is just the cosine transform of $g$ multiplied by $x$. So it suffices to show that the cosine (or Fourier) transform of $g$ is positive, that is, $g$ is positive definite. There is a number of sufficient conditions (e.g. doi.org/10.1017/S0004972700047511), I did not attempt to see if any of them works here, though. Oct 22, 2022 at 8:43
• I don't think any of these sufficient conditions applies here... Oct 22, 2022 at 10:09
• I'm confused here; I would think this could be evaluated by contour integration, closing the contour in the upper half of the complex plane and picking up the poles of the tangent at $(n+1/2)\pi$, $$\tfrac{1}{2}\,{\rm Im}\,\int_{-\infty}^\infty \tanh t\; e^{-\cosh t} e^{ixt}\,dt=\pi\sum_{n=0}^\infty e^{-x(n+1/2)\pi}=\frac{\pi/2}{\sinh(\pi x/2)},$$ but this is mistaken; for small $x$ the integral should be $\propto x$ not $\propto 1/x$... Oct 22, 2022 at 13:28
• @MateuszKwasnicki I attempted to use Tuck's paper, but was unable to see an application. In particular, the kernel here is not even. Using absolute values also makes it not analytic. Ignoring those for a moment, I didn't find a decreasing kernel when shifting the integration up to the singularity, so I don't think that technique would help. I will double check my math. Oct 22, 2022 at 17:14
• @CarloBeenakker Because of the rapid decreasing kernel the Taylor series is valid, hence isn't the function in question an entire function? Oct 22, 2022 at 17:19

1. The reason why the contour integration briefly discussed in Carlo Beenakker's comment is not working here is as follows: For $$t=u+iv$$ with real $$u$$ and $$v$$ such that $$|u|$$ is large and (say) $$\cos v\le-1/2$$, we have $$|\tanh t|\ge\frac{|\sinh2u|}{1+\cosh2u}\approx1$$ and $$|e^{-\cosh t}|=e^{-\Re\cosh t}=e^{-\cosh u\,\cos v}$$, which is very large. So, the modulus of the integrand $$\frac12\,\tanh t\,e^{-\cosh t}\,e^{ixt}$$ of the integral $$$$\int_{-\infty}^\infty \frac12\,\tanh t\,e^{-\cosh t}\,e^{ixt}\,dt$$$$ will be very large for such $$t=u+iv$$, and therefore one cannot control a relevant contour integral.
2. Note that $$\int_5^\infty e^{-\cosh t}\,dt=-\text{Ei}(-\cosh (5))<10^{-34}$$. So, for $$x>0$$, $$$$I(x):=\int_0^\infty \tanh t\,e^{-\cosh t}\,\sin xt\,dt=I_5(x)+\de(x),$$$$ where $$$$I_5(x):=\int_0^5 \tanh t\,e^{-\cosh t}\,\sin xt\,dt$$$$ and $$|\de(x)|<10^{-34}$$.
Below is a Mathematica version of the graph $$\{(x,g(x))\colon0, where $$$$g(x):=\frac{\ln(I_5(x)/x)}{\frac\pi2\,(x+1)}.$$$$ So, it appears that $$$$I(x)=xe^{-(c+\ep(x))x},$$$$ where $$c$$ is a little greater than $$\frac\pi2$$ and $$\ep(x)$$ is oscillating for large $$x>0$$ between $$\ep_*$$ and $$-\ep_*$$ for some $$\ep_*\in(0,c-\frac\pi2)$$; cf. the (incorrect) contour-integration expression of $$I(x)$$ discussed in the mentioned comment. In particular, $$I(x)$$ seems to be exponentially small for large $$x$$, and Mathematica cannot accurately balance the contributions of the positive and negative parts of the integrand of $$I(x)$$ for $$x>24$$.
This apparently exponential decrease of $$I(x)$$ is in contrast with the power decrease of the integral $$$$\int_0^\infty t\,e^{-t}\,\sin xt\,dt=\frac{2 x}{\left(x^2+1\right)^2}$$$$ with the integrand $$t\,e^{-t}\,\sin xt$$ somewhat similar to the integrand $$\tanh t\,e^{-\cosh t}\,\sin xt$$ of $$I(x)$$.