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This example from "Walter A Strauss-Partial differential equations _an introduction-Wiley(2009)" book page 108.

My question is : from where the nonzero coefficient come if $sin(m\pi)=0$

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Nawal
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For $m=0$, you can't take the integral normally since otherwise, you would get a division-by-zero error for $\frac{2}{mx}$. Thus, instead, we have to plus in $m$ before we take the integral:

$$\int_0^l\cos\left(\frac{0\cdot \pi x}{l}\right)dx=\int_0^l1dx=l$$

(Note that I used $\cos 0=1$ in the first step to simplify.)

This leads to a non-zero coefficient for $m=0$.

Noble Mushtak
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