Is my lecturers solution incorret? Integral in relation to a partial differential equation

Question

In this problem we are trying to work out the coefficients $a_n$ in the infinite series for the wave equation on an interval.

We have that $$a_n = \frac{2}{L} \int_0^L x \sin\left(\left(n+\frac{1}{2}\right)\frac{\pi}{L}x\right) $$ Where $n = 0,1,2,\dots$
Now, my lecturer says that $$a_n = \frac{8L}{\pi^2(1+2n)^2}$$ But when I do integration by parts, I find that the answer should be $$a_n = \frac{8L}{\pi^2(1+2n)^2} \sin\left(n+\frac{1}{2}\right)\pi $$

Has my lecturer made a mistake in the notes?

Answer 1

Integrating by parts (differentiating $x$) yields

$$\begin{align*} a_n &= \frac 2 L \left[ x \left( -\frac L \pi \frac{2}{2n+1} \cos \left( \frac{2n+1}{2} \frac \pi L x \right) \right) \right] \Bigg|_0^L \\ &\qquad + \frac 2 L \left( \frac L \pi \frac{2}{2n+1} \right)^2 \sin \left( \frac{2n+1}{2} \frac \pi L x \right) \bigg|_0^L \end{align*}$$

• When $x=0$, the first summand goes away.
• When $x=L$, in the first summand, you have $\cos((2n+1)\pi/2)$, which is $0$ for integers $n$.

Thus the first term is zero.

• When $x=L$, the second summand has $\sin((2n+1)\pi/L)$ which is $(-1)^n$ for integers $n$.
• When $x=0$, then $\sin 0 = 0$ and it vanishes.

So I would conclude you are right and that you can simplify further to

$$a_n = \frac 2 L \left( \frac L \pi \frac{2}{2n+1} \right)^2 (-1)^n = \frac{8L}{\pi^2} \frac{(-1)^n}{(2n+1)^2}$$

It'd be hard to miss the sine term popping up. My guess is your instructor either misremembered some trig identities, or meant to refer to $|a_n|$.