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As part of a boundary value problem, I did the following integral:

$b_n = 2\int^1_0 (30 \sin(\pi x) - 100 (1-x)) \sin (n\pi x) dx$

next step:

$b_n = 2\int^1_0 (30 \sin(\pi x)sin (n\pi x) - (100 (1-x)) \sin (n\pi x) dx$

we can break this up

$ = 60\int^1_0 \sin(\pi x)sin (n\pi x) - 200\int^1_0((1-x)) \sin (n\pi x) dx$

and integrating the two parts, the first half via the product-to-sum identity

$ = 60\int^1_0 \frac{1}{2}[\cos(\pi x- n\pi x) - \cos(\pi x+ n\pi x)]dx - 200\int^1_0((1-x)) \sin (n\pi x) dx$

using inegration by parts on the second half, with $u= 1-x$ and $du = -dx$, $dv = \sin (n\pi x) dx$ and $v = \frac{\cos (n\pi x)}{n \pi}$

$ = 60\int^1_0 \frac{1}{2}[\cos(\pi x- n\pi x) - \cos(\pi x+ n\pi x)]dx - 200[(1-x)(\frac{\cos (n\pi x)}{n \pi})+\int^1_0\frac{\cos (n\pi x)}{n \pi}dx]$

$$ = \left[ 60 \frac{1}{2}[\frac{\sin(\pi x- n\pi x)}{\pi - n\pi } - \frac{\sin(\pi x+ n\pi x)}{\pi + n\pi}] - 200[(1-x)(\frac{\cos (n\pi x)}{n \pi})+\frac{\sin (n\pi x)}{(n \pi)^2}]\right]^1_0$$

And when I do the algebra I get: $$= 30 [\frac{\sin(\pi - n\pi )}{\pi - n\pi } - \frac{\sin(\pi + n\pi )}{\pi + n\pi}] - 200\left(\frac{\sin (n\pi )}{(n \pi)^2}\right) + 200(\frac{1}{n \pi})$$

Here's what I don't get. If n= 1 the part multiplied by 30 goes to 0/0. Does that mean I can count it as 1? That's what the text notes, that it is 0 if $n \neq 1$ and 1 if $n=1$, but I just wanted to be sure I understood this and did the whole integration correctly.


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"If n= 1 the part multiplied by 30 goes to 0/0. Does that mean I can count it as 1?" You certainly can't, whatever your text says! You have to treat n=1 as a special case, and perform the calculation step by step. – TonyK Nov 18 '13 at 19:17
ok, so what is that special case? That is, should i plug in n=1 earlier on (like before I do the integration, when it is still a cosine term) and go from there? Problem is, I still end up with a zero in the denominator... – Jesse Nov 18 '13 at 19:35
Oh wait,, the $\cos(\pi x - \pi x)$ (when n=1) is a 1, and that means it's just x in the integration which is 1... I got it... – Jesse Nov 18 '13 at 19:40
Exactly!${}{}{}$ – TonyK Nov 18 '13 at 19:46

You can justify the step if you consider the parameter-dependent integral

$$I(\alpha) = \int_0^1 \cos (\alpha\pi x)\,dx.$$

For $\alpha \neq 0$, you have

$$\begin{align} I(\alpha) &= \frac{\sin (\alpha\pi x)}{\alpha\pi}\bigl\lvert_0^1\\ &= \frac{\sin (\alpha\pi)}{\alpha\pi}. \end{align}$$

The integrand is continuous and depends continuously on $\alpha$, hence $I(\alpha)$ depends continuously on $\alpha$, and

$$I(0) = \lim_{\alpha\to 0} I(\alpha) = \lim_{\alpha\to 0} \frac{\sin (\alpha\pi)}{\alpha\pi} = 1.$$

But you need to let $\alpha$ assume non-integer values for that. It's at least as easy to just plug in $0$ and compute $\int_0^1 1\,dx$.

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