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NOTE: Please do not provide any sort of a solution to what I am trying to do, as this ia an assessed question. Just let me know whether or not it is a valid use and explain please.

I am trying to show that the following holds: $$\int_{-\infty}^{\infty}\frac{2}{L}\sin{\frac{n\pi x}{L}}\cos{\frac{m\pi x}{L}}dx=0$$ where $n,m\in\mathbb{N}^+$ and $n\neq m$. After integrating by parts I end up with $$\int_{-\infty}^{\infty}\frac{2}{L}\sin{\frac{n\pi x}{L}}\sin{\frac{m\pi x}{L}}dx=0.$$ This is where I decide to use the delta function, but I am unsure whether it is a valid use. I let $$\int_{-\infty}^{\infty}\sin{\frac{n\pi x}{L}}\sin{\frac{m\pi x}{L}}=\delta_{nm}$$ and thus it would follow that, since $m\neq n$, $$\int_{-\infty}^{\infty}\sin{\frac{n\pi x}{L}}\sin{\frac{m\pi x}{L}}=\delta_{nm}=0.$$ Does it make sense to use the delta function in such a way? Or is it being equal to $\delta_{nm}$ a consequence of the fact that it is equal to zero?

ODP
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This does not have anything to do with delta function, as it was written before.

Why do you need a by parts integration? Wouldn't it be more convenient to work with a starting problem, but rewrite it in Euler-like form?

kroniker
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  • Yes, or by using double angle formulae, but I was wondering if it would be possible using delta, which I have now realised you cannot, since while the integral does equal $\delta_{nm}$, this is a consequence of the fact that the integral equals zero, and thus cannot be used in any kind of proof. – ODP Feb 14 '16 at 21:03
  • I was basically trying to show that the stationary wavefunctions for two different quantum mechanical particles are orthogonal, i.e. $\int_{-\infty}^{\infty}dx\phi_n(x)\phi_m(x)=0$ for $n\neq m$, and since this is true, it is also true that $\int_{-\infty}^{\infty}dx\phi_n(x)\phi_m(x)=\delta_{nm}$, which means that it cannot be used in the proof of the orthogonality. – ODP Feb 14 '16 at 21:19
  • I see. Double angle is also an option, but I guess that Euler notation will make all the calculatuion shorter in this case. OK then. – kroniker Feb 14 '16 at 23:55