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The following is one of my graduate analysis past year paper questions.

Question: Evaluate $$\lim_{n\to\infty} \int_0^{\infty} \frac{\ln(n+x)}{n}e^{-x} \cos(x)\, dx.$$ All main steps must be clearly shown.

My attempt:

For each $n\in\mathbb{N}$ and each $x\in \mathbb{R},$ define $$f_n(x) = \frac{\ln(n+x)}{n}e^{-x} \cos(x).$$

By L'Hopital rule, for all $x>0,$ $$\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty}\left(\frac{\ln(n+x)}{n}e^{-x} \cos(x)\right) = e^{-x}\cos(x) \lim_{n\to\infty}\left(\frac{\ln(n+x)}{n}\right) = 0.$$

So the sequence $(f_n)_{n\geq 1}$ converges pointwise.

Observe that for each $x>0,$ we have $\ln(n+x) \leq n+x$ for all $n\in\mathbb{N}.$ It follows that $$|f_n(x)| \leq \left| \frac{\ln(n+x)}{n} \right| \leq \left| \frac{n+x}{n} \right|.$$ Since the function $\left| \frac{n+x}{n} \right|$ is continuous, and hence integrable, by the Dominated Convergence Theorem, we have

$$\lim_n \int_0^{\infty} \frac{\ln(n+x)}{n}e^{-x} \cos(x)\, dx = \int_0^\infty \lim_{n\to\infty}\frac{\ln(n+x)}{n}e^{-x} \cos(x)\, dx = \int_0^\infty 0 \, dx = 0.$$

Is my attempt correct?

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  • $\begingroup$ $(n+x)/n$ is not integrable on $[0,\infty).$ $\endgroup$ – zhw. Aug 25 '17 at 3:56
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Hint: Use the estimate $\ln (n+x) \le \ln (2n)$ on $[0,n].$ Use the estimate $\ln (n+x)\le \ln (2x)$ on $[n,\infty).$

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By the Cauchy-Schwarz inequality, for any $x\geq 1$ we have $$ \log(x)=\int_{1}^{x}\frac{dz}{z}\leq\sqrt{(x-1)\int_{1}^{x}\frac{dz}{z^2}}=\sqrt{x}-\frac{1}{\sqrt{x}}\leq \sqrt{x} $$ hence for any $n\geq 1$ $$ \mathcal{J}(n)=\int_{0}^{+\infty}\log(x+n)e^{-x}\cos(x)\,dx\leq \int_{0}^{+\infty}\sqrt{x+n}\,e^{-x}\left|\cos x\right|\,dx $$ and by the Cauchy-Schwarz inequality again $$ \left|\mathcal{J}(n)\right|\leq\sqrt{\int_{0}^{+\infty}(x+n)e^{-x}\,dx\int_{0}^{+\infty}e^{-x}\cos^2 x\,dx}=\sqrt{\frac{3}{5}(n+1)}$$ hence the given limit is clearly zero.

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  • $\begingroup$ +1 (The upper bound in the second integral of the first line should be $x$ instead of $z$.) $\endgroup$ – Niklas Aug 25 '17 at 9:53

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