How to show that
$$\sum_{n\geq 1} \frac1n \int_0^{\pi/2} \cos(2nx)dx \\ = \int_0^{\pi/2}\sum_{n\geq 1}\frac1n \cos(2nx)dx$$
Classic Riemann and Lebesgue theorems seem not to work here...
Asked
Active
Viewed 38 times
-1
-
Well, the upper, sum is zero as all those integrals are zero... – DonAntonio Feb 17 '23 at 23:32
-
When the upper limit of the sum (N) is finite, the equation is true. Question becomes existence limit as $N\to \infty$. – herb steinberg Feb 17 '23 at 23:44
-
Ok for the left integral which is equal to 0 but the right side? – badinmaths Feb 18 '23 at 18:00
1 Answers
1
Both sides have an explicit closed form. The LHS is zero so you can prove the right side is also zero: take the principal branch of the $\ln(\cdot)$ function, then
\begin{align*} \sum_{n=1}^{\infty} \frac{\cos(2nx)}{n} =&\Re \left(\sum_{n=1}^{\infty} \frac{e^{2xin}}{n}\right)\\ =& \Re \left(-\ln(1-e^{2xi})\right)\\ =& \Re \left(-\ln(1-\cos(2x)-i\sin(2x))\right)\\ =& -\ln\left|(2-2\cos(2x)\right| \\ =& -\ln\left(4\sin^2(x)\right) \end{align*}
So
\begin{align*} \int_{0}^{\frac{\pi}{2}} \sum_{n=1}^{\infty} \frac{\cos(2nx)}{n} dx =& -\int_{0}^{\frac{\pi}{2}}\ln(4\sin^2(x))dx \\ =& -\pi\ln(2)-2\int_{0}^{\frac{\pi}{2}}\ln(\sin(2x))dx \\ =&0 \end{align*}
the last integral is well-known and has the closed form $\displaystyle -\frac{\pi}{2}\ln(2)$
Bertrand87
- 2,171