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

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 ?

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