# Integral Joke

My friend sent me a math joke earlier this year. She was hoping that I would take the bait and attempt the problem, and she won! :)

For those of you who remember advanced calculus techniques, this may be a familiar problem. It is unsolvable using traditional integral calculus techniques.

There are already a lot of solutions online, including solutions computed by calculators, so I won’t be going over the correct solution. Instead, I find it more interesting to show how integration by parts does not solve the problem.

~~I also just really wanted to test out the LaTeX functionality on this website ;')~~

## Integration By Parts

We first start off with the general formula for an integration by parts.

### Integration By Parts Formula

The point of this formula is to break down complicated integrands composed of a product of two functions. In our case, one function would be $sin(x)$, and the other, $\frac{1}{x}$.

Next, we attempt to choose $f(x)$ and $g(x)$ such that it is easy to differentiate $f(x)$ and integrate $g'(x)$. What follows are just math computation steps.

At this point, you probably notice that the second half of the computation in (1), $\int_0^\infty cos(x)ln(x)$, requires another integration by parts operation.

However, notice that if I choose $f(x) = ln(x)$ and $g'(x) = cos(x)$, I’d just be going into circles. If I choose $f(x) = cos(x)$ and $g'(x) = ln(x)$, I would be dealing with further nested integration by parts to compute the integral of the natural log. Therefore, both methods are not good ways of solving this problem.

## Solution Using Laplace Transform - Improper Integrals

I didn’t want to leave you all without the solution, so I found an explanation online. This particular solution uses a Laplace Transformation to solve the integral.

Now, you might be wondering why a transformation is needed. It may be that either the limits go to infinity, or there’s a vertical asymptote which makes the function discontinuous.

We can see an example of this happening at $x=0$:

These integrals can be solved, but they cannot be solved using normal methods initially since we don’t know the finite bounds.
In other words, these are **improper integrals**. If we want to solve them like a regular integral, we need to transform
them into a proper one.

With that being said, enjoy the solution! (~~and maybe see who else you can trick into solving this problem, ha~~)

— AbdullahSalim (@math_phys1) December 19, 2018