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Integral #6

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Show that $$\sum_{k=1}^{N} (-1)^{k} \sin(kx) = - \frac{1}{2} \tan \left(\frac{x}{2} \right) + \frac{(-1)^{N} \sin \Big((N+\frac{1}{2})x \Big)}{2\cos (\frac{x}{2})} \ .$$

Now assuming $ \displaystyle \int_{a}^{b} f(x) \tan \left(\frac{x}{2} \right) \ dx $ converges, argue that

$$ \int_{a}^{b} f(x) \tan \left(\frac{x}{2} \right) \ dx = -2 \sum_{k=1}^{\infty} (-1)^{k} \int_{a}^{b} f(x) \sin (kx ) \ dx \ .$$