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I am having trouble with an integral:
$$\frac2L\int_L^\infty C\sin\left(\frac{n\pi x}{L}\right) dx$$ Where $C$ is just a constant.
I can't see how to do this, despite it apparently being rather straight forward. Something isn't making sense.
$$\frac{2C}L\int_L^\infty \sin\left(\frac{n\pi x}{L}\right)dx$$ $$\frac{2C}L\left[-\frac{L}{n\pi}\cos\left(\frac{n\pi x}{L}\right)\right]_L^\infty$$
But already I can't see how I can compute a divergence $\cos$.
Thank you for listening.
So our integral is:
$$\frac2L\int_L^\infty C\sin\left(\frac{n\pi x}{L}\right) dx$$
Using improper integrals:
$$\lim_{b\to\infty}\frac{2}{L}\int_L^b C\sin\left(\frac{n\pi x}{L}\right) dx$$
$$\lim_{b\to\infty}\frac{2}{L} \left[C\sin\left(\frac{n\pi x}{L}\right)\right]_L^b dx$$
$$\lim_{b\to\infty}\frac{2}{L}\left(C\sin\left(\frac{n\pi b}{L}\right) - C\sin\left(\frac{n\pi L}{L}\right)\right)$$
Since
$$\lim_{b\to\infty}C\sin\left(\frac{n\pi b}{L}\right) = \infty$$
The integral diverges.
The integral is divergent (as your computation shows).
