Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am currently learning analytic number theory using Davenport's Multiplicative Number Theory book, and at some point I believe something silly is happening. I have great faith that I am wrong AND that I am right. Here's the thing.

At some point during the proof of Dirichlet's proof of the theorem about primes in arithmetic progressions, using Poisson's summation formula, one ends up with the integral $$ \int_{-\infty}^{\infty} e^{2\pi i N y^2} \, dy. $$ Here $N$ is an integer. This improper integral should, in practice, evaluate to $\frac{1+i}{2\sqrt N}$. Actually, the reason why we need to compute this is because we know that something we need is equal to $$ \lim_{Y,Z \to \infty} \int_{-Y}^{Z+1} e^{2\pi i N t^2} \, dt, $$ where $Y$ and $Z$ take integer values, and because of some Fourier series argument (which is not the purpose of my question), I know that this limit exists. The way Davenport computes this limit is by evaluating the above integral (not the limit one, the one above it). The way it is computed is by first proving that it converges, and then uses some identity which I don't have problems with. The argument Davenport uses is that for $Y' > Y > 0$, we have $$ \int_Y^{Y'} e^{2\pi i N y^2} \, dy = \frac 12 \int_{Y^2}^{Y'^2} \frac{e^{2\pi i N z}}{\sqrt z} \, dz $$ after the change of variables $z = y^2$, and "this is where I'm stuck, magic happens" : supposedly that "after using the second mean value theorem, or by integration by parts, this should have absolute value $O(\frac 1Y)$ as $Y \to \infty$". How is that? I know both integration by parts/second mean value theorems, but I have no idea how to get there ; naive applications of those two give me no big-oh at all ; for instance, $$ \frac 12 \int_{Y^2}^{{Y'}^2} \frac{e^{2 \pi i N z}}{\sqrt z} \, dz = \frac 12 \left( \left. \frac{e^{2 \pi i Nz}}{2 \pi i N \sqrt z} \right|_{Y^2}^{{Y'}^2} + \frac 1{4\pi i N} \int_{Y^2}^{{Y'}^2} \frac{e^{2\pi i N z}}{(\sqrt z)^3} \, dz \right) $$ The first term is $O(1/Y)$, but the second term, if I use the mean value theorem there's a $Y'$ in the numerator, which can get the thing arbitrary large ; what I want is that the integral from $Y$ to $\infty$ to be bounded for $Y$ large enough, so this is really annoying. I could integrate by parts again, but I would still get a $Y'$ in the numerator.

Another thing that annoys me is that is I write $$ \int_{-\infty}^{\infty} e^{2\pi i N y^2} \, dy = \int_{-\infty}^{\infty} \cos(2\pi N y^2) + i \sin(2 \pi N y^2) \, dy, $$

the real part and imaginary parts both don't seem to converge, since the tail doesn't go to zero, but oscillates very fast? I know that functions can oscillate and still be integrable in the Riemann-limit sense, but still, this looks suspicious.

Any explanations? Anything is welcome... thanks in advance!

share|cite|improve this question

If I understand your question correctly, you want to prove that $$\dfrac1{4 \pi i N} \int_{Y^2}^{Y'^2} \dfrac{e^{2 \pi i Nz}}{z^{3/2}} dz = \mathcal{O} \left( \dfrac1Y\right)$$ where $Y' > Y$. \begin{align} \left \lvert \dfrac1{4 \pi i N} \int_{Y^2}^{Y'^2} \dfrac{e^{2 \pi i Nz}}{z^{3/2}} dz \right \rvert & \leq \dfrac1{4 \pi N} \int_{Y^2}^{Y'^2} \left \lvert \dfrac{e^{2 \pi i Nz}}{z^{3/2}} \right \rvert dz = \dfrac1{4 \pi N} \int_{Y^2}^{Y'^2}\dfrac1{z^{3/2}} dz\\ & = \dfrac1{4 \pi N} \left. \dfrac{z^{-3/2+1}}{-3/2+1} \right \rvert_{Y^2}^{Y'^2}=-\dfrac1{2 \pi N} \left. z^{-1/2} \right \rvert_{Y^2}^{Y'^2}\\ & = -\dfrac1{2 \pi N} \left(Y'^{-1} - Y^{-1} \right) = \dfrac1{2 \pi N} \left( \dfrac1Y - \dfrac1Y' \right)\\ &\leq \dfrac1{2 \pi N Y} = \mathcal{O} \left(\dfrac1Y\right) \end{align}

share|cite|improve this answer
God, I did not even think of this. Thanks for letting me know where I missed! – Patrick Da Silva Jun 18 '12 at 2:41
@PatrickDaSilva For the second part, one way is to go via contour integration to make sense of this integral. Essentially we are integrating along a line between two complex numbers in the complex plane. Since the integrand is analytic, split this into horizontal and vertical segments. Understanding the integrals along these horizontal and vertical segment should hopefully make sense. – user17762 Jun 18 '12 at 3:10

To answer the last part of your question, yes this is one of the unintuitive differences between infinite series and improper integrals. An improper integral can converge because the integrand oscillates quickly, without going to $0$.

Intuitively, the idea is that there isn't much difference between $\int_0^B f(t)\,dt$ and $\int_0^{B+x} f(t)\,dt$ when $B$ is large enough, regardless of $x$: if $x$ is small then it's obvious that $[B,B+x]$ doesn't have much mass, and if $x$ is large then the oscillation creates a lot of cancellation on $[B,B+x]$. Once we can uniformly bound $\left| \int_B^{B+x} f(x)\,dx \right|$, it's not such a leap to make $\lim\limits_{B\to\infty} \int_0^B f(x)\,dx$ converge.

More concretely, if you integrate by parts you get $$\int_0^B\cos(y^2)\,dy = \int_0^B 2y \cos(y^2) \cdot \frac1{2y}\, dy = \sin(y^2)/2y\, \Big|_0^B+ \int_0^B\frac{\sin(y^2)}{2y^2}\, dy,$$ and it is very easy to accept that this converges (of course we also get this from the change of variables in the question).

share|cite|improve this answer
This takes away a lot of magic! Thanks a lot =) – Patrick Da Silva Jun 18 '12 at 3:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.