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I am asked to calculate

$$\lim_{n\rightarrow \infty}\int_0^\infty \frac{e^{\sin(\ln(x))}}{1+\sqrt nx^{42}}dx $$ It looks like it goes to $0$.

I tried

$\int_0^\infty \frac{e^{\sin(\ln(x))}}{1+\sqrt nx^{42}}dx \le \int_0^\infty \frac{e^{1}}{1+\sqrt nx^{42}}dx = e\cdot\int_0^\infty \frac{1}{1+\sqrt nx^{42}}dx = e \cdot \int_0^\infty \frac{1}{1+\sqrt n \cdot (x^{21})^{2}}dx$

First I tried the MCT to get that limes into the integral but MCT cant be used here.

Now I know that $\int\frac{1}{1+x^2} = \arctan$ but I am not really sure on how to use it here.

Any help is appreciated thanks!!

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Substitute $u=n^{1/84}x$ so that $du=n^{1/84}dx$ and the integral equals: \begin{align} \int^\infty_0 \frac{e^{\sin(\ln(x))}}{1+\sqrt[]{n}x^{42}}\, dx = \int^\infty_0\frac{1}{n^{1/84}} \frac{\exp\left( \sin(\ln(un^{-1/84}))\right)}{1+u^{42}}\, du \end{align} You know that $\exp\left( \sin(\ln(un^{-1/84}))\right)\leq e$, hence: \begin{align} \bigg | \frac{1}{n^{1/84}} \frac{\exp\left( \sin(\ln(un^{-1/84}))\right)}{1+u^{42}}\bigg| \leq \frac{e}{1+u^{42}} \end{align} Furthermore $$\int^\infty_0 \frac{e}{1+u^{42}}du<\infty$$ By DCT we have: \begin{align} \lim_{n\to\infty} \int^\infty_0 \frac{e^{\sin(\ln(x))}}{1+\sqrt[]{n}x^{42}}\, dx=\int^\infty_0 \lim_{n\to\infty}\frac{1}{n^{1/84}} \frac{\exp\left( \sin(\ln(un^{-1/84}))\right)}{1+u^{42}}\, du = 0 \end{align}

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  • $\begingroup$ Ahh I didnt even think about substituting. Thank you very much $\endgroup$
    – user391105
    Dec 15 '17 at 19:03
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\begin{align*} \left|\dfrac{e^{\sin(\ln x)}}{1+\sqrt{n}x^{42}}\right|&\leq\left|\dfrac{e^{\sin(\ln x)}}{1+\sqrt{n}x^{42}}\chi_{0< x\leq 1}(x)\right|+\left|\dfrac{e^{\sin(\ln x)}}{1+\sqrt{n}x^{42}}\chi_{x>1}(x)\right|\\ &\leq\dfrac{e}{1+\sqrt{n}x^{42}}\chi_{0< x\leq 1}(x)+\dfrac{e}{1+\sqrt{n}x^{42}}\chi_{x>1}(x)\\ &\leq e\cdot\chi_{0< x\leq 1}(x)+\dfrac{e}{1+x^{42}}\chi_{x>1}(x), \end{align*} and \begin{align*} \int_{0}^{\infty}e\cdot\chi_{0< x\leq 1}(x)+\dfrac{e}{1+x^{42}}\chi_{x>1}(x)dx=e+e\int_{1}^{\infty}\dfrac{1}{1+x^{42}}dx<\infty, \end{align*} so Lebesgue Dominated Convergence Theorem applies.

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    $\begingroup$ Actually one does not have to split the interval $(0,\infty)$, see @robjohn answer. $\endgroup$
    – user284331
    Dec 15 '17 at 19:22
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For each $n,$ the integrand, which is nonnegative, is bounded above on $[0,\infty)$ by $\dfrac{e}{1+\sqrt n x^{42}}.$ Letting $x = n^{-1/84}y$ shows

$$\int_0^\infty \frac{1}{1+\sqrt n x^{42}}\, dx = n^{-1/84}\int_0^\infty \frac{1}{1+y^{42}}\, dy.$$

The last integral is finite and $n^{-1/84}\to 0,$ so our limit is $0.$

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When $x\gt0$, $$ \left|\frac{e^{\sin(\log(x))}}{1+\sqrt{n}x^{42}}\right|\le\frac{e}{1+x^{42}} $$ and $$ \int_0^\infty\frac{e}{1+x^{42}}\,\mathrm{d}x=\frac{e\pi}{42}\csc\left(\frac\pi{42}\right) $$ converges. Furthermore, pointwise, $$ \lim_{n\to\infty}\frac{e^{\sin(\log(x))}}{1+\sqrt{n}x^{42}}=0 $$ Therefore, by Lebesgue Dominated Convergence, we have $$ \lim_{n\to\infty}\int_0^\infty\frac{e^{\sin(\log(x))}}{1+\sqrt{n}x^{42}}\,\mathrm{d}x=\int_0^\infty0\,\mathrm{d}x $$

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  • $\begingroup$ Yes, I just realised that I don't have to split the interval $(0,\infty)$. $\endgroup$
    – user284331
    Dec 15 '17 at 19:12

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