When You're High on Math

As there are so many threes involved and as this is a trigonometric series, it seems only logical to try and work with the expressions for $sin3\theta$ and $cos3\theta$ .

Consider:

$sin3\theta = 3sin\theta -4(sin\theta)^3 = sin\theta \left ( 3-4(sin\theta)^2 \right )$

$=sin\theta \left ( 1+2\left (1-2(sin\theta)^2 \right ) \right )$

$=sin\theta \left ( 1+2cos2\theta \right )$

That's it! We're essentially done because you can use the same relation for the $sin\theta$ in the RHS:

$sin3\theta = sin\theta ( 1+2cos2\theta) =sin\frac{\theta}{3}(1+2cos\frac{2 \theta}{3})( 1+2cos2\theta)$

Keep doing this for every sine term on the RHS and you get: $sin3\theta=\lim_{n\rightarrow \infty}sin\left ( \frac{\theta}{3^n}\right )\prod_{k=0}^{n}\left(1+2 \cos \left(\frac{2\theta}{3^k}\right)\right)$

Divide both sides by $\theta$ and note that $\lim_{n\rightarrow \infty}\frac{sin\left ( \frac{\theta}{3^n}\right )}{\theta}=3^n$

Let $3\theta =x$

From this, we get, $f(x)=\frac{sinx}{x}$

The title for the problem was motivated by the integrals involved.

$\sum_{n=1}^{4} \int_{0}^{\infty}\left ( \frac{sinx}{x} \right )^kdx$

If you are seeing these for the first time and sit down to do them one by one, without the use of contour integration, you really must be high on math!(or maybe meth (just kidding). It's open to debate as to which is the more potent drug.)

$\int_{0}^{\infty}\frac{sinx}{x}dx=\frac{\pi}{2}$

It is a well known result and is called the Dirichlet integral . Follow the link for proofs of the same.

Using this, we can find the other three integrals as well, using integration by parts. It is rather lengthy and I won't post the method here as each of these integrals have already been posted as questions on Brilliant itself.

For example, here .

The final answer evaluates to $\frac{41\pi}{24}$

Thus, $A+B=\boxed{65}$