Revisiting Basel

Some while ago, I showed a slightly dicey proof of the Basel Problem identity, $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$, and invited readers to share other proofs with me.

My old friend Jean Reinaud stepped up to the mark with an exercise from his undergraduate textbook:

The French isn’t that difficult, but just in case:

Determine real numbers $\alpha$ and $\beta$ such that ${\int }_0^{\pi }(\alpha t+\beta t^2)\cos (nt)\mathrm{d}t=\frac{1}{n^2}$, for any integer $n>0$.

Hence deduce $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$

As usual, I suggest having a go at this yourself before reading on. Below here be spoilers.

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Setting up the integral

The first part isn’t too tricky: integrating by parts and noting that $\sin (n\pi )=\sin (0)=0$ for integer $n$ means the integral evaluates to $\frac{(a+2\pi b)\cos (\pi n)-a}{n^2}$.

We need the top to be 1, and we split it into two cases: $n$ even and $n$ odd (since $\cos (\pi n)$ takes on different values in those cases.)

In the odd case, $\cos (\pi n)=1$, so $a+2\pi b-a=1$, giving $b=\frac{1}{2\pi }$.

In the even case, $\cos (\pi n)=-1$, so $-a-2\pi b-a=1$. However, $2\pi b=1$, so $-2a-1=1$, so $a=-1$.

Our integral is ${\int }_0^{\pi }(\frac{t^2}{2\pi }-t)\cos (nt)\mathrm{d}t$.

The tricky bit: the infinite sum

So, to get our result, presumably we just need to sum all of those integrals from 1 to $\mathrm{\infty }$, and boom ¹! Out should pop our result.

The brackety bit doesn’t have anything to do with $n$, so we can write our summed integral as:

${\int }_0^{\pi }(\frac{t^2}{2\pi }-t)\sum _{n=1}^{\mathrm{\infty }}\cos (nt)\mathrm{d}t$.

Evaluating that sum, we hit a problem: it doesn’t converge for any value of $t$. For example, for $t=\pi$, $\cos (nt)$ oscillates between $-1$ and $1$ - and that sum is undefined. Even worse, for $t=0$, $\cos (nt)$ is always 1, so we end up with a sum that goes to infinity. It’s a disaster!

But.

We can engage in a bit of jiggery-pokery with the partial sums, and show that they converge to the right thing.

Let’s start with $\sum _{n=0}^N\cos (nt)$. Depending on how interested you are, you can:

• a) Express each cosine as $\frac{e^{int}+e^{-int}}{2}$, add up the resulting geometric series and play about until you get the result;
• b) Look it up on Wolfram Alpha; or
• c) Look it up on Wikipedia.

All three, thankfully, give the same result: $\sum _{n=0}^N\cos (nt)=-\frac{1}{2}+\frac{\sin \left((N+\frac{1}{2})t\right)}{2\sin \left(\frac{t}{2}\right)}$.

Well, that’s a mess. However, it does reduce a pretty substantial sum to something that only depends on $N$ and $t$.

Evaluating that mess

But how to evaluate that, combined with the quadratic $t$ factor? We’re after:

${\int }_0^{\pi }(t-\frac{t^2}{2\pi })[-\frac{1}{2}+\frac{\sin \left((N+\frac{1}{2})t\right)}{2\sin \left(\frac{t}{2}\right)}]\mathrm{d}t$

Let’s split that hairy square bracket into two parts: the $-\frac{1}{2}$ bit will be easy to integrate, but ${\int }_0^{\pi }(t-\frac{t^2}{2\pi })\left[\frac{\sin \left((N+\frac{1}{2})t\right)}{2\sin \left(\frac{t}{2}\right)}\right]\mathrm{d}t$… less so. So much less so, I suggest calling it $I_N$.

Jean’s textbook gives us a hint: $\cot (t)<\frac{1}{t}<\mathrm{cosec}\left(t\right)$ over the domain we care about. In particular, $\left[\frac{\sin \left((N+\frac{1}{2})t\right)}{2\sin \left(\frac{t}{2}\right)}\right]>\left[\frac{\sin \left((N+\frac{1}{2})t\right)}{t}\right]$

That gives us a lower bound for $I_N$:

$I_N={\int }_0^{\pi }(t-\frac{t^2}{2\pi })\left[\frac{\sin \left((N+\frac{1}{2})t\right)}{2\sin \left(\frac{t}{2}\right)}\right]\mathrm{d}t>{\int }_0^{\pi }(t-\frac{t^2}{2\pi })\left[\frac{\sin \left((N+\frac{1}{2})t\right)}{t}\right]\mathrm{d}t$

Pushing the $t$ in the denominator back into the quadratic factor, we get:

$I_N>{\int }_0^{\pi }(1-\frac{t}{2\pi })\sin \left((N+\frac{1}{2})t\right)\mathrm{d}t.$

Stick it in the parts formula, turn the crank, note that $\sin (N\pi )=0$ for integer $N$ and out pops $I_N>\frac{-2\cos (N\pi )+2(1+2N)\pi }{{(1+2N)}^2\pi }$.

This is good: the lower bound goes to zero as $N$ gets large.

Great. However, that’s just a lower bound. What about an upper bound?

Can we express it another way?

Of course we can.

Let’s come back to here:

$\frac{\sin \left((N+\frac{1}{2})t\right)}{2\sin \left(\frac{t}{2}\right)}$

… and expand the sine in the numerator. We get $\sin \left((N+\frac{1}{2})t\right)=\sin (Nt)\cos \left(\frac{1}{2}t\right)+\cos (Nt)\sin \left(\frac{1}{2}t\right)$.

Of course, there’s a $\sin \left(\frac{1}{2}t\right)$ in the denominator as well, so we can simplify to get $\sin \left(Nt\right)\cot \left(\frac{1}{2}t\right)+\cos (Nt)$. Suddenly the future is brighter.

… at least momentarily. $I_N={\int }_0^{\pi }(t-\frac{t^2}{2\pi })(\sin \left(Nt\right)\cot \left(\frac{1}{2}t\right)+\cos (Nt))\mathrm{d}t$… is not Wolfram-friendly. However, Jean’s textbook comes to the rescue again: we know that $\cot \left(\frac{1}{2}t\right)<\frac{2}{t}$.

So, $I_N<{\int }_0^{\pi }(t-\frac{t^2}{2\pi })(\frac{2\sin \left(Nt\right)}{t}+\cos (Nt))\mathrm{d}t$. If I’ve done my sums right, that’s $\frac{1-2N+N\cos (N\pi )}{N^2}$ - which gives us an upper bound that goes to 0 as $N$ gets large.

So: $I_N$ is no bigger than an upper bound that goes to 0 in the limit, and no smaller than a lower bound that also goes to zero in the limit. By the squeeze theorem, as $N$ gets large, $I_N$ goes to 0.

Phew.

Back to the integral

${\int }_0^{\pi }(t-\frac{t^2}{2\pi })[-\frac{1}{2}+\frac{\sin \left((N+\frac{1}{2})t\right)}{2\sin \left(\frac{t}{2}\right)}]\mathrm{d}t$

Everything after the $-\frac{1}{2}+$ can be ignored - we just proved that bit of the integral goes to 0 in the limit.

Meanwhile, $-\frac{1}{2}{\int }_0^{\pi }(t-\frac{t^2}{2\pi })\mathrm{d}t=\frac{{\pi }^2}{6}$, as required $◼$.

* If you’ve enjoyed this post, and want to do something kind: please sign up to be an organ donor. Without someone like you saying “I want to help someone live after I die,” Jean would not be with us today. I’m on the register - I’d be delighted if you joined me.

Footnotes:

1. Avoiding surprise factorial