How do you find #\int _ { 0} ^ { x } \frac { \sin ( t ) } { t } d t#?

1 Answer
Jun 4, 2017

Your integral cannot be expressed in terms of elementary functions.

Explanation:

Unfortunately, your integral cannot be computed using elementary functions.

A series solution can be computed in order to compute the integral to an arbitrary amount of accuracy, though.

#int_{0}^{x}(\sin(t))/(t) "d"t = \int_{0}^{x} (\sum_{n=1}^{\infty}(-1)^n(t^(2n+1))/((2n+1)!))/(t) "d"t#
#int_{0}^{x}(\sin(t))/(t) "d"t = int_{0}^{x} \sum_{n=1}^{\infty}(-1)^n(t^(2n))/((2n+1)!) "d"t#

#int_{0}^{x}(\sin(t))/(t) "d"t = \sum_{n=1}^{\infty}(-1)^n(x^(2n))/((2n+1)(2n+1)!) #.

Interestingly though, there is a closed form for the indefinite integral across the entire domain which I won't prove here but you can look up online.

#int_{-\infty}^{\infty}(sin(x))/x=\pi#