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I got curious about a Mathoverflow question and so I read about the so called explicit formula about the zeta function in Davenport's book in analytic number theory. Everything looks good to me except the following discontinuous integral:


where $c > 0$, and $y>0$, and the improper integral is understood as


of the integrand. The claim is it is equal to 0 if $y<1$, $\dfrac{1}{2}$ if $y=1$, and 1 if $y>1$. When I tried to work this out it boils down to the following real valued integral, namely


which equal to $\dfrac{2\pi}{e^A}$ if $A>0$, $\pi$ if $A=0$, and $0$ if $A<0$. As can be checked by Wolfram Alpha here.

I took complex analysis years ago so I don't remember how this is evaluated, can anybody here help me out? ($A=0$ is trivial.)

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(+1) for using computeralgebra. – vesszabo Oct 30 '12 at 19:17
This is covered in Murty's book as well, in the beginning of chapter 4: (If you're on a university campus, you should be able to get a free download.) – Andrew Oct 30 '12 at 19:21
$\int_{-\infty}^{\infty}\frac{\cos(Av)}{1+v^2}$ can be found here Example II. – vesszabo Oct 30 '12 at 19:25
up vote 5 down vote accepted

What you have written is called Perron's formula and is a very important first step in analytical number theory: $$ \frac1{2\pi i} \int_{c - i\infty}^{c+i\infty} \frac{y^s}{s} \, ds = \begin{cases} 1 & y > 1 \\ \frac12 & y = 1 \\ 0 & y < 1 \end{cases} $$ for $y > 0$ and $c > 0$ (not just $c>1)$.

Let us first analyze what the integral means. First, what do the bounds mean? This is a contour integral, and we evaluate it on the line $c + it$. Does this integral make sense as such? The integral does not converge absolutely, and hence we have to be careful. The integral should be interpreted as $$\int_{c-i\infty}^{c+i\infty} = \lim_{T \to \infty} \int_{c-iT}^{c+iT}$$

Let's first do a simple case, when $y = 1$. In this case, we can compute the integral: $$ \frac1{2\pi i} \int_{c - iT}^{c + iT} \frac1s \, ds = \frac1{2\pi} \int_{-T}^T \frac{dt}{c + it} = \frac1{2\pi} \int_0^T \left({\frac{1}{c + it} + \frac1{c - it}} \right)\, dt = \frac1{2\pi} \int_0^T \frac{2c}{c^2 + t^2} \, dt, $$ which is some arctangent that we can easily compute.

Note that this doesn't depend on $c$. Why should we expect this to happen? If we integrated along some other line, we can shift from one contour to the other. When we do this, we don't cross any singularities because we assume that $c > 0$, so therefore the integral should be the same along any contour.

Before proving it, below is what we will expect.

For $y < 1$, we have that $y^c \to 0$ as $c \to +\infty$. Then we move the line of integration to the right, so the integrand becomes small, so the integral is $0$. For $y > 1$, we have $y^c \to 0$ as $c \to -\infty$, so we move the line of integration to the left. But the difference here is that we cross a singularity at $0$, and the residue of the singularity is $y^0 = 1$.


We will work with the integral $$\frac1{2\pi i} \int_{c - iT}^{c + iT} y^s \frac{ds}{s}$$

First, consider the case $y < 1$; we want to move the contour to the right. We can write $$\int_{c-iT}^{c+iT} = \int_{c - iT}^{d-iT} + \int_{d-iT}^{d+iT} - \int_{c+iT}^{d+iT}$$ We just have to estimate the integrals on these three other sides. First, look at the horizontal integral (and write $s = \sigma - iT$) $$\left \vert{\int_{c-iT}^{d-iT} \frac{y^s}s \, ds} \right \vert \leq \int_c^d \frac{y^\sigma}{T} \, d\sigma \leq \frac1T \int_c^\infty y^\sigma \, d\sigma = \frac{y^c}{|\log y| T}$$ The same bound holds for $\int_{c+iT}^{d+iT}$. The last thing to think about is $$\left \vert{\frac1{2 \pi i} \int_{d-iT}^{d+iT} \frac{y^s}{s} \, ds } \right \vert \leq C y^d \int_{-T}^T \frac{dt}{1 + |t|} \leq C y^d \log T$$ Now, let $d \to \infty$. This term goes to zero, and we already had good estimates for the horizontal terms. So we have that if $0 < y < 1$ and $c > 0$ then $$\left \vert{ \frac1{2 \pi i} \int_{c-iT}^{c+iT} \frac{y^s}{s} \, ds} \right \vert \leq \frac{y^c}{\pi T |\log y|}$$ Taking the limit as $T \to \infty$ yields that $$\lim_{T \to \infty} \frac1{2\pi i} \int_{c-iT}^{c+iT} \frac{y^s}s \, ds = 0$$ So not only have we proved this, we've even proved this in a more quantitative way.

Let's think a bit more about this $\log y$ term. It is natural to expect to get a term like this, because we know that there is a discontinuity at $y = 1$, and we'll run into problems there.

Now, we evaluate the integral for $y > 1$; we want to move the contour to the left. So we can write $$\frac1{2\pi i} \int_{c-iT}^{c+iT} = \frac1{2\pi i} \left({\int_{c-iT}^{-d - iT} + \int_{-d-iT}^{-d+iT} + \int_{-d+iT}^{c+iT}} \right) + 1 $$ where the "$+1$" comes since we have a pole at $s = 0$ and by shifting the contour we pick up that pole. Now, we have the same type of argument as before, estimating each integral separately. We have $$\left \vert{\int_{-d-iT}^{-d+iT} \frac{y^s}s \, ds} \right \vert \leq C y^{-d} \int_{-T}^T \frac{dt}{1 + |t|} \leq C y^{-d}\log T \to 0$$ as $d \to \infty$.

For the horizontal integrals, we have precisely the same bounds as before: they are bounded by $$ \int_{-d}^c \frac{y^\sigma}{T} \, d\sigma \leq \frac{y^c}{T |\log y|}.$$

So therefore when $y > 1$ and $c > 0$, we have $$\left \vert{\frac1{2\pi i} \int_{c-iT}^{c+iT} \frac{y^s}s \, ds - 1} \right \vert \leq \frac{y^c}{\pi T |\log y|}$$ so therefore $$\lim_{T \to \infty} \frac1{2\pi i} \int_{c-iT}^{c+iT} \frac{y^s}{s} \, ds = 1.$$

Hence, we have

$$ \frac1{2\pi i} \int_{c - i\infty}^{c+i\infty} \frac{y^s}{s} \, ds = \begin{cases} 1 & y > 1 \\ \frac12 & y = 1 \\ 0 & y < 1 \end{cases} $$ for $y > 0$ and $c > 0$.

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Thanks! I just realized how amazing this is, since it allows you to pick up information from any Dirichlet series. I wasn't able to appreciate such formula when I began to learn complex analysis, they used to look very bizarre to me. – h__ Oct 30 '12 at 20:13

This was supposed to be a response to the comment made by hyh but it was way too long.

Of course it works. Let $A > 0$ and take the same contour as the one used to calculate $\int_{-\infty}^\infty \frac{\cos(Ax)}{1+x^2}dx$. Then $$ \int_\mathcal{C} \frac{z e^{iAz}}{1+z^2}dz = \int_{-a}^a \frac{x e^{iAx}}{1+x^2}dx + \int_0^\pi \frac{a e^{i\theta} e^{iA a e^{i\theta}}}{1 + a^2 e^{2i\theta}}aie^{i\theta}d\theta. $$

The second integral can be bounded by $$ \int_0^\pi \frac{a^2 e^{-A a\sin\theta}}{a^2 - 1}d\theta \le \int_0^\pi \frac{a^2 e^{-A a\theta\left(1-\frac{\theta}{\pi}\right)}}{a^2 - 1}d\theta = \frac{2\sqrt{\pi} a^{3/2}e^{-\frac{Aa\pi}{4}}}{\sqrt{A}(a^2-1)}\int_0^\frac{\sqrt{Aa\pi}}{2}e^{u^2}du.$$ Then $$\int_0^\pi \frac{a e^{i\theta} e^{iA a e^{i\theta}}}{1 + a^2 e^{2i\theta}}aie^{i\theta}d\theta \le \frac{2\sqrt{\pi} a^{3/2}D_+\big(\tfrac{\sqrt{Aa\pi}}{2}\big)}{\sqrt{A}(a^2-1)} \to 0 \mbox{ if }a \to \infty, $$ given that the Dawson Integral $D_+(x)$ is bounded.

Finally, you can evaluate the residue $$ \int_\mathcal{C} \frac{z e^{iAz}}{1+z^2}dz = \int_\mathcal{C} \frac{z e^{iAz}}{(z+i)(z-i)} dz = 2 \pi i \frac{z e^{iAz}}{z+i}\bigg|_{z=i} = i \pi e^{-A}, $$ and then $$ \int_{-\infty}^\infty \frac{x \sin(A x)}{1+x^2}dx = \Im\left(\int_{-\infty}^\infty \frac{x e^{iAx}}{1+x^2}dx\right)= \pi e^{-A}. $$

Note: In the case $A < 0$, you need to take the arc enclosing $-i$ instead, and the residue evaluates to $-i\pi e^{A}$, which concludes the proof.

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Great. Thanks for letting me know. You seem really good at this! – h__ Oct 31 '12 at 1:08
@hyh Not really. It's standart first course complex analysis. I'm sure you can find a prettier proof in Marsden's Complex Analysis book. – Pragabhava Oct 31 '12 at 1:23
I think your proof is pretty enough, using only some basic estimates. When I was in college I thought only physicists know how to integrate such thing! BTW I like the remark at the end. – h__ Oct 31 '12 at 2:20
Yes, it is "the" standard method can be found in complex analysis books. If $A<0$ then using that the function $\sin$ is odd we obtain the result at once (without any more complex calculation). In the book Bak and Newman, Complex Analysis Ch 11, p.132-133, there are formulae (with proof) for $\int_{-\infty}^{\infty}R(x)\cos(x)dx$ and $\int_{-\infty}^{\infty}R(x)\sin(x)dx$ where $R(x)=P(x)/Q(x)$, $\deg Q>\deg P$ and $Q(x)\neq0$ (except perhaps at a zero of $\cos x$ and $\sin x$. Introducing new variables in hyh's integrals the results follows. – vesszabo Oct 31 '12 at 17:04

Since Marvis does not answer your question (although the answer is absolutely correct) I give a non-complete answer. Hopefully someone else will be able to justify it. I do not want to copy here the calculations from wikipedia. (See my comment above.) So it is enough to calculate $\int_{-\infty}^{\infty}\frac{v\sin(Av)}{1+v^2}dv$. Denote$f(A):=\int_{-\infty}^{\infty}\frac{\cos(Av)}{1+v^2}dv.$ If we could show that the order of integration and differentiation is interchangeable then we obtain $$ f'(A)=-\int_{-\infty}^{\infty}\frac{v\sin(Av)}{1+v^2}dv=\left(\pi e^{-A} \right)'=-\pi e^{-A}, $$ which gives $$ \int_{-\infty}^{\infty}\frac{v\sin(Av)}{1+v^2}dv=\pi e^{-A}. $$ So try to prove that this derivation is correct.

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You don't need to do that, just do the same calculation of wikipedia but with $\int_{-\infty}^\infty \frac{x e^{i A x}}{1+x^2}dx$ instead. – Pragabhava Oct 30 '12 at 20:27
@Pragabhava Probably :-) you are right. However, could you prove my "conjecture"? – vesszabo Oct 30 '12 at 20:33
What @Pragabhava said is not quite good. You want to use a contour, which is the boundary of half a disk in this case. So you need some bound of the integral on the disk. The one he suggested doesn't give that bound. – h__ Oct 30 '12 at 22:18
@hyh It works the same. See my answer for details. – Pragabhava Oct 31 '12 at 1:03

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