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Calculate

$\lim_{n \rightarrow \infty} \int_{[0, \infty]} e^{-\sqrt x}$ $\cos$ ($x^2 \over n$) $d\lambda^1(x)$.

My attempt:

First, we want to use the Dominated Convergence Theorem. We define

$f_n(x) := e^{-\sqrt x}$ $\cos$($x^2 \over n$)

and note that this sequence of functions is integrable.

Furthermore, we define $f := \lim_{n \rightarrow \infty} e^{-\sqrt x}$ $\cos$ ($x^2 \over n$) $= e^{-\sqrt x}$.

Since $|e^{-\sqrt x}$ $\cos$($x^2 \over n$)| $\le e^{-\sqrt x}$, we set

$F := f$, which shall be our majorant.

Now we have everything that we need in order to apply the Dominated Convergence Theorem. Thus, the above integral is identical to

$\int_{[0, \infty]} e^{-\sqrt x} d\lambda(x)$.

Since $[0, \infty] = \lim_{m \rightarrow \infty} [0, m]$, we are allowed to write this as

$\lim_{m \rightarrow \infty} \int_{[0, m]} e^{-\sqrt x} d\lambda(x)$ and treat is as a Riemann-integral:

$\lim_{m \rightarrow \infty} \int_0^m e^{-\sqrt x} dx$.

Integrating delivers

$[-2(\sqrt x + 1) e^{-\sqrt x}]_0^m = -2(\sqrt m + 1) e^{-\sqrt m} + 2$, and by taking the limit $m \rightarrow \infty$, the expression $-2(\sqrt m + 1) e^{-\sqrt m}$ converges to $0$, thus,

$\lim_{m \rightarrow \infty} \int_0^m e^{-\sqrt x} dx = 2.$

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    Your limite is $1$ – 2017-02-01
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    Looks good to me. Where is the question though??? – 2017-02-01
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    alternativly, taylor $\cos(x^2/n)$ and integrate termwise... – 2017-02-01
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    Ah, sorry, my question was simply whether the solution is correct or not. @Hamou: Are you sure? Do I have to keep the minus here although the expression vanishes completely? – 2017-02-01
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    @Borol let me clarify: everything looks fine up until the Riemann Integral, which evaluates to $2$ according to Mathematica/WA, and not to $-1$. I presume you made a basic calculation mistake because $e^{-\sqrt{x}} $ is strictly positive, so the integral should be strictly positive. – 2017-02-01
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    Found my mistake, your calculations are right! – 2017-02-01

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