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$$\int_0^a \sin (\frac{n\pi x}{a}) \delta(x-a/2)\sin (\frac{m\pi x}{a})dx$$

For the integral above, I cannot combine the two sine functions easily. But can I do this:

$$=\sin (\frac{n\pi}{2})\sin (\frac{m\pi}{2})$$

? Basically I am treating both sines multiplied together as one function of $x$ and applying the property of dirac delta to this big function. I have another idea to, but not sure which of these two ideas is correct:

$$=\frac{1}{2}\int_0^a \delta(x-a/2) (\cos(\frac{n\pi x}{a}- \frac{m\pi x}{a})-\cos(\frac{n\pi x}{a} + \frac{m\pi x}{a}))dx$$

$$=\frac{1}{2} (\cos(\frac{n\pi }{2}- \frac{m\pi }{2})-\cos(\frac{n\pi }{2} + \frac{m\pi }{2}))$$

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    The definition is $\int_b^c f(x) \delta(x-a)dx = \begin{cases} f(a) \text{ if } a \in (b,c)\\0 \text{ if } a \not \in [b,c] \end{cases}$ whenever $f$ is continuous at $a$ – 2017-02-22

2 Answers 2

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They're both entirely correct (as long as $0

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    Wait actually, $a$ is not defined. So then how do I do this?? All I know about $a$ is that it is a real number greater than $0$. – 2017-02-22
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    If a isn't in the interval $[0, 1]$, then the integrand is 0 instead, for the reason that user1952009 gave in the comments. – 2017-02-22
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    That doesn't make sense. The interval is $[0,a]$. Of course $a/2$ is within that interval. You said something different - you specified that the interval for $a$, which is different. – 2017-02-22
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    If $a$ isn't inside $[0, 1]$, then $a/2$ isn't inside $[0, 2]$, and so the integral becomes zero. – 2017-02-22
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    Ohhhhhhhhh I see the confusion. I made a mistake in my question. I will edit. Sorry about that. The interval should be $[0,a]$ – 2017-02-22
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Both work: the product of two functions is a function, so there's no difficulty in doing it the first way. And you will notice, by using the prosthaphaeresis formula the other way, that the answers are equal.

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    Sorry, made a mistake. See edited question. Had the interval of integration wrong. – 2017-02-22