Integral Sum 90

Calculus Level 4

Compute

∫ 0 1 ( ∑ n = 2 ∞ cos ⁡ ( π n x ) n ) d x \int_0^1 \left(\sum _{n=2}^{\infty } \frac{\cos (\pi n x)}{n}\right) \, dx


The answer is 0.

Christopher Criscitiello by Christopher Criscitiello , 25, USA

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1 solution

First Last
Jan 1, 2018

Moving the integral to the inside: ∫ 0 1 ∑ n = 2 ∞ cos ⁡ ( π n x ) n d x = ∑ n = 2 ∞ ∫ 0 1 cos ⁡ ( π n x ) n d x = ∑ n = 2 ∞ sin ⁡ ( n x ) n 2 sin ⁡ ( n x ) is always 0 for integer n \displaystyle\int_0^1\sum_{n=2}^\infty\frac{\cos{(\pi n x)}}{n}dx=\sum_{n=2}^\infty\int_0^1\frac{\cos{(\pi n x)}}{n}dx=\sum_{n=2}^\infty\frac{\sin{(nx)}}{n^2}\quad\sin{(nx)}\text{ is always 0 for integer n}

∑ n = 2 ∞ sin ⁡ ( n x ) n 2 = 0 \displaystyle\sum_{n=2}^\infty\frac{\sin{(nx)}}{n^2}=0

2 Helpful 0 Interesting 1 Brilliant 0 Confused

The sum doesn't converge uniformly so how do you justify the change of sum and integral?

Markus Reibnegger - 3 years, 4 months ago

∫ 0 1 ( ∑ n = 2 ∞ C o s ( π n x ) n ) d x \int_0^1(\sum_{n=2}^{\infty}\frac {Cos(\pi nx)}{n})dx = l n ( S i n x ) = − l n ( 2 ) − ∑ n = 1 ∞ C o s ( 2 n x ) n ln(Sinx)=-ln(2)-\sum_{n=1}^{\infty}\frac {Cos (2nx)}{n} = ∫ 0 1 [ ∑ n = 2 ∞ C o s ( n π x n ] d x \int_0^1[\sum_{n=2}^{\infty}\frac {Cos(n\pi x}{n}]dx = ∫ 0 1 [ − l n ( 2 ) − l n ( S i n π 2 x ) − C o s ( π x ) ] \int_0^1 [-ln (2)-ln (Sin\frac{\pi}{2}x)-Cos(\pi x)] = − l n ( 2 ) − 2 π ∫ 0 π 2 l n ( S i n x ) d x -ln (2)-\frac {2}{\pi}\int_0^{\frac {\pi}{2}}ln(Sinx)dx Now using Clausen function C l 2 ( z ) Cl_2 (z) C l 2 ( z ) : = − ∫ 0 z l n ( 2 S i n t 2 ) d t Cl_2(z):=-\int_0^z ln(2Sin\frac {t}{2})dt = ∑ n = 1 ∞ S i n ( n z ) n 2 \sum_{n=1}^{\infty}\frac {Sin (nz)}{n^2} Integral can expressed as − 1 2 C l 2 ( π ) − π 2 l n ( 2 ) -\frac {1}{2}Cl_2(\pi)-\frac {\pi}{2}ln (2) . We can notice that the integral, C l 2 ( m , π ) = 0 Cl_2 (m,\pi)=0 where, ∀ m ∈ Z \forall m\in\mathbb {Z} due to the series representation.

= l n ( 2 ) 2 π ∫ 0 π 2 l n ( S i n x ) d x = − l n ( 2 ) − 2 π [ − 1 2 C l 2 ( π ) − π 2 l n ( 2 ) ] = 0 ln (2)\frac{2}{\pi}\int_0^{\frac {\pi}{2}}ln (Sinx)dx=-ln (2)-\frac {2}{\pi}[-\frac {1}{2}Cl_2 (\pi)-\frac {\pi}{2}ln (2)]=0

Pawan Goyal - 2 years, 1 month ago

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