(i)$$\frac{1}{ \theta^2 - \pi^2} + \frac{1}{ \theta^2- 2^2\pi^2} + \frac{1}{ \theta- 3^2\pi^2 }... $$ (ii)$$\frac{1}{1^2 + x^2} + \frac{1}{2^2 +x^2} + \frac{1}{3^2 + x^2}... $$ These are the problems given in "Differential Calculus for Beginners"" by Joseph Edwards. I am beginner and have no idea how to approach such problems. The phrase 'differentiate logarithmically' is confusing me.
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Factors of what? I don't see any products. – saulspatz Nov 05 '20 at 18:36
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Where are $\sin$ and $\cos$? – Raffaele Nov 05 '20 at 18:41
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https://math.stackexchange.com/questions/3763155/logarithmic-differentiation-of-expressions-of-sin-theta-and-cos-theta This link contains exact picture of the text book from which the question is taken. – Ranjeet Bahadur Nov 05 '20 at 18:43
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I suspect that "the expressions for $\sin\theta$ and $\cos\theta$ in factors" refers to the infinite product representations of $\sin$ and $\cos$. – John Barber Nov 05 '20 at 19:15
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The phrase "the expressions for $\sin\theta$ and $\cos\theta$ in factors" seems to refer to the infinite product representations of those functions.
For (i):
\begin{align*} \sin\theta &= \theta \; \prod_{n=1}^{\infty}\left(1 - \frac{\theta^2}{n^2 \pi^2}\right)\\ \frac{d}{d\theta}\ln(\sin\theta) &= \frac{d}{d\theta}\left[\ln\theta \;+\; \sum_{n=1}^{\infty}\ln\left(1-\frac{\theta^2}{n^2\pi^2}\right)\right]\\ &= \frac{1}{\theta} \;+\; 2\theta \sum_{n=1}^{\infty}\frac{1}{\theta^2 - n^2\pi^2} \end{align*} From there you can solve for the sum in question, to yield: $$ \sum_{n=1}^{\infty}\frac{1}{\theta^2 - n^2\pi^2} \;=\; \frac{1}{2\theta}\left(\cot\theta \;-\; \frac{1}{\theta}\right) $$
For (ii), the simplest thing to do is to use the above result with $\theta\rightarrow i \pi x$.
John Barber
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