I am trying to compute the following limit: $$ \lim_{n\to \infty} \int_0^\infty \frac{x^{n-2}}{1+x^n} \cos(\pi n x) \,dx . $$ This is a problem in an old analysis qualifying exam.

Let $f_n(x) = \frac{x^{n-2}}{1+x^n} \cos(\pi n x)$. Since $$ \lim_{n\to\infty} f_n(x) = \begin{cases} 0 & x \neq 1 \\ 1/2 & x=1 , \end{cases} $$ if I could show that $|f_n| \leq g$ for some $g \in L^1$, then the Dominated Convergence Theorem would imply that the limit in question is zero.

I'm having trouble finding such a $g$ and have started doubting my approach. Any ideas?

  • 1
    $\begingroup$ Try considering $\int_0^1$ and $\int_1^\infty$ separately. $\endgroup$ – Antonio Vargas May 3 '14 at 1:57

The $-2$ in the exponent of the nominator screams "Use $\frac{1}{x^2}$, use $\frac{1}{x^2}$!".

And indeed, for $x \geq 1$, $$ \frac{x^{n-2}}{1 + x^n} = \frac{1}{x^{2-n} + x^2} \leq \frac{1}{x^2} \text{.} $$

For $0 \leq x \leq 1$, we just need that the integrand ins bounded, and bounded it is because $$ \frac{x^{n-2}}{1 + x^n} \leq \frac{1}{1+ x^n} \leq \frac{1}{2} \text{.} $$

Out of pur vanity we double that bound because discontinuity just isn't pretty, and use $$ g(x) = \begin{cases} 1 &\text{if $x \leq 1$} \\ \tfrac{1}{x^2} &\text{if $x > 1$.} \end{cases} $$

  • $\begingroup$ Very nice answer! +1 $\endgroup$ – Bennett Gardiner May 3 '14 at 2:31

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