Newtonian Pedantics

I’m a fan of the Epic Rap Battles of History. In a fairly recent video, science educator Bill Nye is pitted against a giant of physics, Isaac Newton.

Newton’s parting words are

“And I will leave you with a page
from a book I wrote at half your age to rebut:
the integral sec y dy
from zero to one sixth of pi
is log to base e
of the square root of three
times the sixty-fourth power of what?”

That is, Newton wants Nye to solve the equation

\displaystyle \int_0^{\frac{\pi}{6}} \sec y\,dy=\ln\sqrt{3} \,x^{64}

Here’s the answer.

Let’s call the left-hand I:

\displaystyle I=\int_0^{\frac{\pi}{6}} \sec y\,dy

There are two ways of finding the indefinite integral. The first is to guess (a perfectly valid way of solving an equation) and the second is to use substitution to reduce to it a more easily integrable, recognisable form. We’re going to do the latter.

The secant function is defined as being the reciprocal of the cosine function:

\displaystyle I=\int_0^{\frac{\pi}{6}} \frac{1}{\cos y}\,dy

The trick is to multiply the top and bottom of the integrand by cosine:

\displaystyle I=\int_0^{\frac{\pi}{6}}\frac{\cos y}{\cos^2 y}\,dy

The denominator can be rewritten using the identity

\cos^2 y+\sin^2 y=1

\displaystyle I=\int_0^{\frac{\pi}{6}}\frac{\cos y}{1-\sin^2 y}dy

Now we can use the substitution

\varphi=\sin y

The virtue of this substitution is that the differential d\varphi naturally appears in the integrand’s numerator:

d\varphi=\cos y\,dy

Remember the limits of integration must also be changed. The lower limit stays the same, because the sine of zero is zero, but the upper limit changes to \sin(\frac{\pi}{6})=\frac{1}{2}.

\displaystyle I=\int_0^\frac{1}{2}\frac{d\varphi}{1-\varphi^2}

Here we use the method of partial fractions to split up the integrand into two fractions:

\displaystyle I=\frac{1}{2}\int_0^\frac{1}{2}\Bigg(\frac{1}{1+\varphi}+\frac{1}{1-\varphi}\Bigg)d\varphi

So we’ve reduced the integrand to (relatively) friendly rational functions. We integrate:

\displaystyle I=\frac{1}{2}\Big[\ln(1+\varphi)-\ln(1-\varphi)\Big]_0^\frac{1}{2}

\displaystyle I=\frac{1}{2}\Big[\ln\Big(\frac{1+\varphi}{1-\varphi}\Big)\Big]_0^\frac{1}{2}

Now it’s just a case of substituting in the limits of the integration:

\displaystyle I=\frac{1}{2}\Big[\ln\Big(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\Big)-\ln\Big(\frac{1+0}{1-0}\Big)\Big]

\displaystyle I=\frac{1}{2}[\ln(3)-\ln(1)]

The natural logarithm of one is zero (we’re asking, e raised to what power is 1?), so that term vanishes:

\displaystyle I=\frac{1}{2}\ln 3

Finally, we bring the half ‘inside’ the logarithm


because the number to which to raise e to get 3 is twice that you raise e to to get the square root of 3, if you see what I mean.

Now we substitute this expression back into the original equation for x we’re trying to solve:





Fundamental theorem of algebra alert. This equation does not have an answer; it has many answers, sixty-four of them to be exact! Two are pure real, two are pure imaginary, and the remainder are complex. To deduce this, we write  the number 1 in complex form:

1=e^{2n\pi i}

where n is any integer. We can then take the sixty-fourth root of the equation to give

\displaystyle x_n=\exp\bigg(\frac{n\pi i}{32}\bigg)

I’ve stuck a subscript n on the answer so we can keep track of which root we’re talking about. Substitute in any integer value of n and you will get a perfectly valid solution to this equation. Let’s look at a few of them. We have the naïve solution


and the slightly less naive solution


We also have the solutions



Then we have a load of weird ones like

\displaystyle x_{17}=\cos\bigg(\frac{17\pi}{32}\bigg)+i\sin\bigg(\frac{17\pi}{32}\bigg)

and so on. Here are the solutions to the equation plotted in the complex plane:


As you may know, Nye does not get a chance to rebut. Instead, Neil deGrasse Tyson appears (naturally) and claims

“By the way,
the answer to your little calculation is i,
as in, I put the swag back in science.”

So he should really have said ‘an answer’.

This kind of stuff is important. Peer review and that.

They’ll thank me one day.

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