What is the integral from 1 to infinity of sin(x^2) ?

The integral from 1 to infinity of sin(x^2) is (sqrt(pi))/(2sqrt(2))-sqrt(pi/2)\S(sqrt(2/pi)*1)

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integral from 1 to infinity of sin(x^2)

\int_{1}^{\infty} sin (x^{2}) d x

\frac{π}{2 2} - \frac{π}{2} § (\frac{2}{π} \cdot 1)

Solution steps

\int_{1}^{\infty} sin (x^{2}) d x

Compute the indefinite integral:

\int sin (x^{2}) d x = \frac{π}{2} § (\frac{2}{π} x) + C

Compute the boundaries:

\int_{1}^{\infty} sin (x^{2}) d x = \frac{π}{2 2} - \frac{π}{2} § (\frac{2}{π} \cdot 1)

= \frac{π}{2 2} - \frac{π}{2} § (\frac{2}{π} \cdot 1)

What is the integral from 1 to infinity of sin(x^2) ?
The integral from 1 to infinity of sin(x^2) is (sqrt(pi))/(2sqrt(2))-sqrt(pi/2)\S(sqrt(2/pi)*1)