A Sum Of Zeroes

Calculus Level 1

∫ 0 π cos ⁡ x d x = 0 ∫ π 2 π cos ⁡ x d x = 0 ∫ 2 π 3 π cos ⁡ x d x = 0 ∫ 3 π 4 π cos ⁡ x d x = 0 ⋮ \begin{aligned} \displaystyle \int_0^{\pi } \cos x \, dx &= & 0 \\ \displaystyle \int_{\pi}^{2\pi} \cos x \, dx &= & 0 \\ \displaystyle \int_{2\pi}^{3\pi} \cos x \, dx &= & 0 \\ \displaystyle \int_{3\pi}^{4\pi} \cos x \, dx &= & 0 \\ &\vdots & \end{aligned}

Because all the equation above are true, is the following equation true as well?

∫ 0 ∞ cos ⁡ x d x = ∫ 0 π cos ⁡ x d x + ∫ π 2 π cos ⁡ x d x + ∫ 2 π 3 π cos ⁡ x d x + ⋯ = 0 + 0 + 0 + ⋯ = 0 . \begin{aligned} \int_0^\infty \cos x \, dx &= &\int_0^\pi \cos x \, dx + \int_{\pi}^{2\pi} \cos x \, dx + \int_{2\pi}^{3\pi} \cos x \, dx + \cdots \\ &=& 0+ 0+0+ \cdots \\ &=& 0 \; . \end{aligned}

True False

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Pi Han Goh by Pi Han Goh , 30, Malaysia

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3 solutions

Shourya Pandey
Apr 9, 2016

The given relations only tell us that

∫ m π k π cos ⁡ ( x ) = 0 \displaystyle \int_{m\pi}^{k\pi} {\cos (x)} =0 for all finite non-negative integers m and k . But that doesn't mean ∫ 0 ∞ c o s x = 0 \int_0^\infty {cosx} =0 .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

0 isn't a natural number.

Aditya Kumar - 5 years, 2 months ago

beautiful solution..this is why we say that 0*infinity is an indeterminate value

guido barta - 5 years, 1 month ago

This isn't the proof that ∫ 0 ∞ cos ⁡ ( x ) ≠ 0 \displaystyle \int_0^\infty {\cos(x)} \neq 0 . You'll have to take cos ⁡ ( x ) = e i x + e − i x 2 \cos(x)=\frac{e^{ix}+e^{-ix}}{2} and show that it diverges.

Aditya Kumar - 5 years, 2 months ago

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We can also use methods other than the euler's formula, for example,
consider the series ∫ 0 a 1 cos ⁡ x d x , ∫ 0 a 2 cos ⁡ x d x , ⋯ \displaystyle\int_0^{a_1}\cos x \; \mathrm{dx},\int_0^{a_2}\cos x \; \mathrm{dx},\cdots
If a n = n π a_n=n\pi , the series converge to 0 0
If a n = ( n + 1 2 ) π a_n=(n+\frac 12)\pi , the series converge to 1 1
Since they does not match, the series diverges.

展豪 張 - 5 years, 1 month ago

The function F ( t ) = ∫ 0 t c o s ( x ) d x F(t) = \int^t_0 cos(x) dx doesn't approach a limit when t goes towards infinity. The value of the funxtion is oscillating.

Looking at the macroscopic change in the value of the function does not tell us about the limit

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Yeah exactly my thought. The sin(x) as antiderivative doesn't converge towards a limit, it diverges/oscillates.

Marko Kohler - 5 years, 2 months ago
Noel Lo
Jun 9, 2016

Infinity is not clearly defined. For us to get zero as the answer, both the upper and lower limits need to be clearly defined in this case.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Infinity IS clearly defined.

Pi Han Goh - 5 years ago

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