In this course, we've learned skills in five key areas:

  • Limits
  • Differentiation
  • Integration
  • ODEs
  • Series

Some of the things we can do are pretty impressive. However, there are many simple-seeming questions in single-variable calculus that show us just how much we have left to learn.


Why is the standard Gaussian a probability density function? In other words, why is

Why is it that with all our methods, we cannot evaluate this integral easily? Let's try...


This integral is easy with a little bit of multivariable calculus.


Recall that we began this class with the definition of the exponential function and then, to obtain series for and , we invoked Euler's formula:

Why is this true? We certainly could substitute in our favorite Taylor series and verify that it is true, but wouldn't it be better to have a principled reason for why this is so? Let's try...

Let . Then by that very familiar differential equation. Now, name the real and imaginary parts of by and respectively. Then , and . On the other hand, multiplying by gives

Therefore, becomes , and so by equating the real and imaginary parts in this equation we get the system

This is a system of differential equations which is easy to solve with some multivariable calculus, but for now we are stuck. We can observe that and provides a solution, but we cannot say it is the only solution without more tools.


This system of ODEs is easy to solve with a little bit of multivariable calculus.


On several occasions, we have referenced the famous series

Why is this true? Could we use discrete calculus to derive it? I don't think so...but here is proof using Taylor series and integration.

Let . Consider the integral

On the other hand, the integral in terms of is

One can show that the Taylor series for is given by

Plugging this in gives

One can find with careful integration by parts and induction that the inner integral evaluates to

If we encountered this integral earlier in the course, we would hit it with a trigonometric substitution (and , which changes the integral to

This integral can be found inductively using the following reduction formula:

Use integration by parts. Let and . Then and . It follows that

Now, solving for the integral (by adding to both sides of the above equation and dividing by ) gives the desired result.

Applying the reduction formula in the situation at hand gives

since the first quantity evaluates to 0. Now, induction gives the result.

Plugging this in cancels almost everything, leaving

This is the sum of the odd reciprocals squared. Giving names to these various sums:

Note that ( has the odd terms, and has the even terms). Further,

Substituting, we find that , and so . Since as shown above, it follows that , as desired.


This series is easy to evaluate with a little bit of multivariable calculus. Well, actually, no: it's not easy, but it is a bit simpler. In the end, some math problems are hard.