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The following problem has appeared in 2013 January qualifying exam in Purdue University, which is publicly available here.

Problem 3. Let $\{a_k\}$ be sequence of positive numbers such that $a_n\to\infty$ as $n\to\infty$. Prove that the following limit exists $$ \lim_{k\to\infty}\int_{0}^{\infty} \frac{e^{-x}\cos(x)}{a_kx^2 + \frac{1}{a_k}} dx $$ and find it.

I have hardly come across to limits of sequences that involve definite integrals (in my undergraduate education so far), so this problem just seems insurmountable at the first glance. I would appreciate any hints.

One of the things that comes to mind is to use limit comparison test. For example, we can evaluate integrals such as $$\int_{0}^{\infty} e^{-x}\cos(x)=\frac{1}{2}$$ But for that we would have to bound the integrand somehow. One tempting thing is to interchange the integral and the limit, which would tell us that integrand is zero in the limit, but I highly doubt this is allowed here.

Looking forward to hear your thoughts.

P.S. I am not sure how to make the title informative for this post. Feel free to edit as you see fit.

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  • $\begingroup$ what does "converging to infinity" mean? $\endgroup$ – Ron Gordon Jun 26 '13 at 7:48
  • $\begingroup$ @RonGordon: I have edited accordingly. $\endgroup$ – Prism Jun 26 '13 at 7:52
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Substitute $y = a_k \cdot x$ in the integral. You obtain

$$\int\limits_0^\infty \frac{e^{-y/a_k}\cos (y/a_k)}{y^2 + 1}\, dy$$

And now you can apply the dominated convergence theorem, the integrand converges to $\frac{1}{1+y^2}$ pointwise and is dominated by the limit, hence the integrals tend towards

$$\int\limits_0^\infty \frac{dy}{1+y^2} = \frac{\pi}{2}.$$

If the dominated convergence theorem is not available since one is dealing with Riemann integrals, one can also obtain the result from splitting the substituted integral at strategic points.

Fix $z > 1$ arbitrarily. For all $k$ such that $a_k > z^3$, we can split the integral at $z$, write $I(k,z) = \int_0^z \frac{e^{-y/a_k}\cos (y/a_k)}{1+y^2}\,dy$ and $II(k,z) = \int_z^\infty \frac{e^{-y/a_k}\cos (y/a_k)}{1+y^2}\,dy$, and can estimate

$$\lvert II(k,z)\rvert \leqslant \int\limits_z^\infty \bigg\lvert \frac{e^{-y/a_k}\cos (y/a_k)}{1+y^2}\bigg\rvert\,dy \leqslant e^{-z/a_k}\int\limits_z^\infty \frac{dy}{1+y^2} < \frac{\pi}{2} - \arctan z$$

for the second part, and

$$\lvert\arctan z - I(k,z)\rvert = \Biggl\lvert\int\limits_0^z \frac{1 - e^{-y/a_k}\cos (y/a_k)}{1+y^2}\,dy \Biggr\rvert \leqslant \int\limits_0^z \frac{\lvert 1 - e^{-y/a_k}\cos (y/a_k)\rvert}{1+y^2}\,dy$$

for the first part.

Now, we assumed that $a_k > z^3$, so $0 \leqslant y/a_k \leqslant z/a_k < 1/z^2$ for $I(k,z)$, so $\cos (y/a_k) \geqslant 1 - \frac{1}{2z^4}$ and $e^{-y/a_k} \geqslant e^{-1/z^2} > 1 - \frac{1}{z^2}$. We can hence estimate the numerator

$$\lvert 1 - e^{-y/a_k}cos (y/a_k) \rvert \leqslant 1 - \biggl(1 - \frac{1}{z^2}\biggr)\biggl(1 - \frac{1}{2z^4}\biggr) = \frac{1}{z^2} + \frac{1}{2z^4} - \frac{1}{2z^6} < \frac{2}{z^2}.$$

Thus we obtain $\lvert\arctan z - I(k,z)\rvert \leqslant z\cdot\frac{2}{z^2} = \frac{2}{z}$.

Altogether

$$\biggl\lvert\frac{\pi}{2} - \bigl(I(k,z) + II(k,z)\bigr) \biggr\rvert \leqslant \biggl(\frac{\pi}{2} - \arctan z\biggr) + \lvert \arctan z - I(k,z)\rvert + \lvert II(k,z)\rvert \leqslant \frac{2}{z} + \pi - 2\arctan z.$$

The last quantity obviously tends towards $0$ for $z \to \infty$.

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  • $\begingroup$ Thanks for your answer! I guess it's time for me to learn the dominated convergence theorem, of which I know nothing about. Looking at Wikipedia, it says it is used in the context of Lebesgue integrals. Do you have a reference to the result you used above, in the Riemann integral setting? $\endgroup$ – Prism Jun 26 '13 at 8:01
  • $\begingroup$ Uh, Riemann integral is a bit harder. I'm almost sure you can prove the dominated convergence theorem there too under a few additional conditions [you need to know that the limit is nice (locally Riemann integrable), which is the case here, and possibly a few other things]. But here, we can also use another technique to obtain the result, I'm going to add that. $\endgroup$ – Daniel Fischer Jun 26 '13 at 8:09
  • $\begingroup$ Thank you Daniel. (+1) And nice substitution there! I will look forward to seeing the "another technique" you mention in your comment. $\endgroup$ – Prism Jun 26 '13 at 8:14
  • $\begingroup$ @Prism Whew, that took a bit longer to write down than I thought. But it doesn't use any theory beyond high school (well, at least what high schools taught in my day). $\endgroup$ – Daniel Fischer Jun 26 '13 at 8:44
  • $\begingroup$ Brilliant strategy, and yes it is nice and elementary. I will come back in the morning to work through the estimates more carefully, and then accept the answer :) $\endgroup$ – Prism Jun 26 '13 at 8:55

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