What is the integral from 0 to x of (cos(t)-1)/t ?

The integral from 0 to x of (cos(t)-1)/t is diverges

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integral from 0 to x of (cos(t)-1)/t

\int_{0}^{x} \frac{cos ( t ) - 1}{t} d t

diverges

Solution steps

\int_{0}^{x} \frac{cos ( t ) - 1}{t} d t

Expand

\frac{cos ( t ) - 1}{t} : \frac{cos ( t )}{t} - \frac{1}{t}

= \int_{0}^{x} \frac{cos ( t )}{t} - \frac{1}{t} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{x} \frac{cos ( t )}{t} d t - \int_{0}^{x} \frac{1}{t} d t

\int_{0}^{x} \frac{cos ( t )}{t} d t = Ci (x)

\int_{0}^{x} \frac{1}{t} d t =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{0}^{1.3} \frac{1}{6} x d x

What is the integral from 0 to x of (cos(t)-1)/t ?
The integral from 0 to x of (cos(t)-1)/t is diverges