# Analyze a function defined in terms of an integral

Here is a question that really has puzzled me for quite a while. I happened to see this function defined in terms of an integral $$f(x):=\int_0^{\pi/2}\frac{2e^{x+e^x\cos y}}{1+\left(e^{e^x\cos y}\right)^2}dy.$$I want to analyze the behavior of the function when $$x \rightarrow \infty$$.

The strange thing is that when I used Mathematica to plot the function, the graph indicates that $$\lim_{x\rightarrow \infty} f(x)=0$$. However, it is easy to see that $$\liminf_{x\rightarrow \infty}f(x) \ge \frac{\pi}{4}$$, since $$\int_0^{\pi/2}\frac{2e^{x+e^x\cos y}}{1+\left(e^{e^x\cos y}\right)^2} \, dy \ge \int_0^{\pi/2}\frac{2e^{x+e^x\cos y}\sin y}{1+\left(e^{e^x\cos y}\right)^2}\, dy\\ =-\Big(\tan^{-1}\left(e^{e^x \cos y}\right)\Big)\Big|_{0}^{\pi/2}\\=\tan^{-1}\left(e^{e^x}\right)-\pi/4$$

Now I have two questions:

First, why the result from Mathematica is different from what I obtained?

Second, does $$\lim_{x\rightarrow \infty} f(x)$$ exist?

Maybe this question is not so suitable for mathoverflow, since it is just a calculus problem. However, I just feel so confused about the contradiction of numerical result and math. I want to understand the reason behind this situation. Any comments are really appreciated. Thank you very much.

Below is the code and picure I got from Mathematica.... • you probably made a coding error in Mathematica, when I plot it the limit is about 1.57 – Carlo Beenakker Jun 14 at 6:58
• You missed $2$ in antiderivative, so $\liminf$ is at least $\frac{\pi}{2}$ (which coincides with the number Carlo wrote). I bet that it won't be hard to prove that the the loss in multiplying by $\sin (y)$ is negligible as $x\to \infty$ and so the limit is actually $\frac{\pi}{2}$. – Aleksei Kulikov Jun 14 at 7:18
• @CarloBeenakker, thank you very much for the comment... As you can see from my updates, my code should be fine, but when I tried to plot the the graph for $x \in [0,20]$, something went wrong... – student Jun 14 at 13:31
• @AlekseiKulikov, exactly! Thank you very much! – student Jun 14 at 13:31

Mathematica seems to be plotting the function just fine... If we look a bit at the integrand, it's clear that most of the mass is around $$y = \pi/2$$ as $$x$$ increases which should let us introduce a $$\sin y$$ term and use the antiderative you've already found. We can try to cut the integral at $$\pi/2 - 1/x$$.

Let $$I = \int_0^{\pi/2 - 1/x} 2 e^x \frac{e^{e^x \cos y}}{1 + e^{2 e^x \cos y}} dy + \int_{\pi/2 - 1/x}^{\pi/2} 2 e^x \frac{e^{e^x \cos y}}{1 + e^{2 e^x \cos y}} dy = I_0 + I_1$$

Notice that $$f(u) = e^u / (1 + e^{2u})$$ is a decreasing function of $$u$$ and thus that $$I_0 < (\pi/2 - 1/x) 2 e^x \frac{e^{e^x \cos (\pi/2-1/x)}}{1 + e^{2 e^x \cos (\pi/2-1/x)}}$$

When $$x \rightarrow \infty$$ the $$\cos (\pi/2 - 1/x)$$ behaves as $$1/x$$ and the logistic function $$1-\sigma(u)$$ behaves as $$e^{-u}$$, so the right term behaves as $$\pi e^{x - e^x /x}$$ which converges to $$0$$.

For $$I_1$$, we note that if $$y \in [\pi/2-1/x,\pi/2]$$, $$1 - \frac{1}{2x^2} < \sin y \leq 1$$

$$I_1 \left(1-\frac{1}{2x^2}\right)< \int_{\pi/2-x}^{\pi/2} 2 e^x \frac{e^{e^x \cos y}\sin y}{1 + e^{2 e^x \cos y}} \leq I_1$$

The middle term converges to $$\pi /2$$ and thus so does $$I_1$$ and so does $$I$$.

• Thank you very much! By the way, I got the same picture for $x$ up to 10, but when I plotted the graph for $x \in [0,20]$, it gives me the limit going to $0$, as you can see from my updates.... – student Jun 14 at 13:26
• The support of the mass becomes too small and the numerical integration misses it. If you do a change of variable which blow up the region around pi/2 it'll be more stable. – Arthur B Jun 14 at 13:44