What is the integral from 0 to t of (1-e^u)/u ?

The integral from 0 to t of (1-e^u)/u is diverges

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integral from 0 to t of (1-e^u)/u

\int_{0}^{t} \frac{1 - e ^{u}}{u} d u

diverges

Solution steps

\int_{0}^{t} \frac{1 - e ^{u}}{u} d u

Expand

\frac{1 - e ^{u}}{u} : \frac{1}{u} - \frac{e ^{u}}{u}

= \int_{0}^{t} \frac{1}{u} - \frac{e ^{u}}{u} d u

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}^{t} \frac{1}{u} d u - \int_{0}^{t} \frac{e ^{u}}{u} d u

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to t of (1-e^u)/u ?
The integral from 0 to t of (1-e^u)/u is diverges