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Let $\displaystyle I(n,\alpha)=\int_0^\pi \sin(n\theta)e^{\alpha \theta} d\theta$. Show that $\displaystyle I(n,\alpha)=\frac{n}{n^2+\alpha^2}(1-(-1)^ne^{\alpha \pi})$.

This is done by parts, but I can't get the expression to equal the RHS.

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What have you done already? It may just be a simplification problem, which makes the question a lot easier to answer... – apnorton Mar 10 '13 at 17:15
@anorton I haven't done anything really. – bbr4in Mar 10 '13 at 21:50

If you are not forced to use integration by parts, you can use the fact that sin is the imaginary part of an exponential:

$$\sin{n \theta} = \Im{[e^{i n \theta}]}$$


$$\begin{align}\int_0^{\pi} d\theta \: \sin{n \theta} e^{\alpha \theta} &= \Im{\left[ \int_0^{\pi} d\theta \:e^{(\alpha+i n) \theta} \right]}\\ &= \Im{\left[ \frac{e^{(\alpha+i n)\pi}-1}{\alpha+i n}\right]}\\ &= \frac{-n}{\alpha^2+n^2} [(-1)^n e^{\alpha n}-1]\end{align}$$

The stated result follows.

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I don't understand how you've got from the second line to the third line which begins to look like the result needed. – bbr4in Mar 11 '13 at 1:49
The numerator is real, as I assume $\alpha$ is real. That leaves taking the imaginary part of $1/(\alpha+i n)$, found by multiplying by $(\alpha-i n)/(\alpha-i n)$. The result follows. – Ron Gordon Mar 11 '13 at 1:53

By parts it's not bad, defining $J(n,\alpha)=\int_0^{\pi}cos(n \theta)e^{\alpha \theta}.$ Applying parts twice gives back an expression involving the original $I(n,\alpha)$. Any reasonable calc text will have at least a specific case of this integral solved by parts, meaning that it may solve e.g. $I(3,2)$. Then you can just look at such an example and watch where the 3 and 2 would be changed, if they were symbols $n$ and $\alpha$ instead of specific numbers.

NOTE: When parts is applied to a definite integral, in the formula $\int u dv=uv-\int v du$ one has to evaluate the $uv$ term at the endpoints of the definite integral (and subtract). So in going through the general case, you'll have to use that $\sin(0)=\sin(n\pi)=0$ and that $\cos(0)=1$ while $\cos(n\pi)=(-1)^n.$ This is the source of the $(-1)^n$ appearing in the final answer.

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Note: I adapted this from Arturo's answer at: Integration by parts

Also, please forgive me for not using your variable names as I found them confusing (easy to modify the entire result by just changing them to your names).

Note: we are going to save the integration limits to the end for ease.

So, doing IBP twice yileds:

$$\int e^{ax}\sin(bx)dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\left(\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).$$

Multiplying out you get $$\int e^{ax}\sin(bx)\,dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a^2}e^{ax}\cos(bx) - \frac{b^2}{a^2}\int e^{ax}\cos(bx)\,dx.$$ At this point, you should move that last integral on the right hand side to the left hand side and "leave oout" the constant of integration on the right (because we'll use limits at the end as noted earlier).

Moving the last integral to the left hand side, you get $$\left(1 + \frac{b^2}{a^2}\right)\int e^{ax}\sin(bx)\,dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a^2}e^{ax}\cos(bx).$$

Now, $\displaystyle 1 + \frac{b^2}{a^2} = \frac{a^2+b^2}{a^2}$, and we want to clear this from the left side, so multiply both sides by the reciprocal, that is by, $\displaystyle \frac{a^2}{a^2+b^2}$.

If you do that, from $$\frac{a^2+b^2}{a^2}\int e^{ax}\sin(bx)\,dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a^2}e^{ax}\cos(bx),$$ multiplying both sides by $\frac{a^2}{a^2+b^2}$, we get: $$\tag 1 \int e^{ax}\sin(bx)\,dx = \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx).$$

Now, we just need to deal with the RHS of $(1)$ with the limits from $\Pi$ to $0$.

So we have:

$\displaystyle \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx)$ evaluated at $\pi$, yields: $\displaystyle \frac{a}{a^2+b^2} e^{a \pi}(a \sin b \pi - b \cos b \pi)$

Of course, $\sin b \pi= 0$ for all $b$ (odd or even) and $\cos b \pi = \pm 1$ (depending on parity of b, that is $-1$ if $b$ is odd and is $+1$ if b is even, so we will account for this by using $(-1)^b$).

$\displaystyle \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx)$ evaluated at $0$, yields: $\frac{a}{a^2+b^2} (0 - b).$

Subtracting these expressions (because we are doing over limits $\pi$ and $0$), yields:

$\displaystyle \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx)|{_0^\pi} = \frac{b}{a^2+b^2}\left(1 - (-1)^b e^{a \pi}\right)$.

The desired result.


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Excellent details, and nice credit and link to Arturo! Keep up the good work, dear friend! – amWhy Apr 21 '13 at 0:34
@amWhy: Thanks - it just looked like a fun problem and I had to give it go - since no one answered the way he asked. Regards – Amzoti Apr 21 '13 at 0:41

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