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the exponential function being increasing we have $| \exp((\cos^n x) -x)| \leq \exp(1 -x) \in L^1([0,+\infty[) $

so $x \to \exp((\cos^n x) -x)$ is Riemann absolutely convergent therefore

$l = \lim_{n \to \infty} \int_{\mathbb{R_{+}}} \exp((\cos x^n) -x) d\lambda(x) =\lim_{n \to \infty} \int_{0}^{+\infty} \exp((\cos^n x) -x) dx $

by the dominated convergence theorem :

$l = \int_{0}^{+\infty} \lim_{n \to \infty} \exp((\cos^n x) -x) dx$

I don't know how to deal with this limit, as $x$ is in $\mathbb{R_{+}}$ I can't even use a taylor expression around $0$

any hints ?

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For $x \in \mathbb R_+ \setminus\{k\pi + \pi/2 \ ; \ k \in \mathbb N\}$ you have:

$$\exp((\cos^n x) -x) \to e^{-x}$$ as $n \to \infty$ and

$$0 \le \exp((\cos^n x) -x) \le e \cdot e^{-x}$$ for all $x \in \mathbb R_+$.

As $\{k\pi + \pi/2 \ ; \ k \in \mathbb N\}$ is a null set (for Lebesgue measure) and $\int_{\mathbb R_+} e^{-x} \ dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that

$$\lim_{n \to \infty} \int_{\mathbb{R_{+}}} \exp((\cos^n x) -x) d\lambda(x) = \int_{\mathbb R_+} e^{-x} \ dx$$

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Notice that as $n \to \infty$ all values of $\cos^n(x)$ go to zero except the exact points at which $\cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $\exp(-x)$.

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