What is the integral from 0 to infinity of |sin(x)| ?

The integral from 0 to infinity of |sin(x)| is diverges

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integral from 0 to infinity of |sin(x)|

\int_{0}^{\infty} ∣ sin (x) ∣ d x

diverges

Solution steps

\int_{0}^{\infty} ∣ sin (x) ∣ d x

Eliminate Absolutes

= \int_{0}^{π} sin (x) d x + \int_{π}^{2 π} - sin (x) d x + \int_{2 π}^{3 π} sin (x) d x + \int_{3 π}^{4 π} - sin (x) d x + \int_{4 π}^{5 π} sin (x) d x + \int_{5 π}^{6 π} - sin (x) d x + \int_{6 π}^{7 π} sin (x) d x + \int_{7 π}^{8 π} - sin (x) d x + \int_{8 π}^{9 π} sin (x) d x + \int_{9 π}^{\infty} - sin (x) d x

\int_{0}^{π} sin (x) d x = 2

= 2 + \int_{π}^{2 π} - sin (x) d x + \int_{2 π}^{3 π} sin (x) d x + \int_{3 π}^{4 π} - sin (x) d x + \int_{4 π}^{5 π} sin (x) d x + \int_{5 π}^{6 π} - sin (x) d x + \int_{6 π}^{7 π} sin (x) d x + \int_{7 π}^{8 π} - sin (x) d x + \int_{8 π}^{9 π} sin (x) d x + \int_{9 π}^{\infty} - sin (x) d x

\int_{π}^{2 π} - sin (x) d x = 2

= 2 + 2 + \int_{2 π}^{3 π} sin (x) d x + \int_{3 π}^{4 π} - sin (x) d x + \int_{4 π}^{5 π} sin (x) d x + \int_{5 π}^{6 π} - sin (x) d x + \int_{6 π}^{7 π} sin (x) d x + \int_{7 π}^{8 π} - sin (x) d x + \int_{8 π}^{9 π} sin (x) d x + \int_{9 π}^{\infty} - sin (x) d x

\int_{2 π}^{3 π} sin (x) d x = 2

= 2 + 2 + 2 + \int_{3 π}^{4 π} - sin (x) d x + \int_{4 π}^{5 π} sin (x) d x + \int_{5 π}^{6 π} - sin (x) d x + \int_{6 π}^{7 π} sin (x) d x + \int_{7 π}^{8 π} - sin (x) d x + \int_{8 π}^{9 π} sin (x) d x + \int_{9 π}^{\infty} - sin (x) d x

\int_{3 π}^{4 π} - sin (x) d x = 2

= 2 + 2 + 2 + 2 + \int_{4 π}^{5 π} sin (x) d x + \int_{5 π}^{6 π} - sin (x) d x + \int_{6 π}^{7 π} sin (x) d x + \int_{7 π}^{8 π} - sin (x) d x + \int_{8 π}^{9 π} sin (x) d x + \int_{9 π}^{\infty} - sin (x) d x

\int_{4 π}^{5 π} sin (x) d x = 2

= 2 + 2 + 2 + 2 + 2 + \int_{5 π}^{6 π} - sin (x) d x + \int_{6 π}^{7 π} sin (x) d x + \int_{7 π}^{8 π} - sin (x) d x + \int_{8 π}^{9 π} sin (x) d x + \int_{9 π}^{\infty} - sin (x) d x

\int_{5 π}^{6 π} - sin (x) d x = 2

= 2 + 2 + 2 + 2 + 2 + 2 + \int_{6 π}^{7 π} sin (x) d x + \int_{7 π}^{8 π} - sin (x) d x + \int_{8 π}^{9 π} sin (x) d x + \int_{9 π}^{\infty} - sin (x) d x

\int_{6 π}^{7 π} sin (x) d x = 2

= 2 + 2 + 2 + 2 + 2 + 2 + 2 + \int_{7 π}^{8 π} - sin (x) d x + \int_{8 π}^{9 π} sin (x) d x + \int_{9 π}^{\infty} - sin (x) d x

\int_{7 π}^{8 π} - sin (x) d x = 2

= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + \int_{8 π}^{9 π} sin (x) d x + \int_{9 π}^{\infty} - sin (x) d x

\int_{8 π}^{9 π} sin (x) d x = 2

= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + \int_{9 π}^{\infty} - sin (x) d x

\int_{9 π}^{\infty} - sin (x) d x =

diverges

= diverges

\int_{- 5}^{5} - 25 - x^{2} d x

\int_{0}^{1} 7 x (x^{2} - 1)^{5} d x

\int_{1}^{3} (2 x^{2}) d x

\int_{1}^{2} 3 x^{2} + 4 x d x

\int_{- 1}^{1} e^{3 x^{2}} d x

What is the integral from 0 to infinity of |sin(x)| ?
The integral from 0 to infinity of |sin(x)| is diverges