Question

41-46 use the comparison theorem to determine whether the given integral is convergent or divergent.

\(\int_1^\infty {\frac{{2 + {e^{ - s}}}}{x}} dx\)

Short answer

The given integral diverges.

Step-by-step solution

Solve the value of the integral\(\int_1^\infty  {\frac{{2 + {e^{ - s}}}}{x}} dx\)

The given integral is \(\int_1^\infty {\frac{{2 + {e^{ - s}}}}{x}} dx\).

It is known that \(0 < {e^{ - x}}\) for all \(x \in R\).

Then,

\(0 < {e^{ - x}}\)

Add\(2\)to both sides,

\(2 < 2 + {e^{ - x}}\)

Divide by\(x\)on both sides,

\(\frac{2}{x} < \frac{{2 + {e^{ - x}}}}{x}\)

Solve the value of the\(\int_1 g (x)\)

Comparison Theorem:

Assume that\(f\)and\(g\)are continuous functions with\(f(x) \ge g(x) \ge 0\)for\(x \ge a\).

(a) If\(\int_a^\infty f (x)dx\)is convergent, then\(\int_a^\infty f (x)dx\)is convergent.

(b) If\(\int_a^\infty f (x)dx\)is divergent, then\(\int_a^\infty f (x)dx\)is divergent

Let \(f(x) = \frac{{2 + {e^{ - x}}}}{x}\)and \(g(x) = \frac{2}{x}\)

Now, by the above definitions, obtain the value of the integral \(\int_1 g (x)\) as shown below:

\(\begin{aligned}{l}\int_1^\infty g (x)dx &= \int_1^\infty {\frac{2}{x}} dx\\\int_1^\infty g (x)dx &= \mathop {\lim }\limits_{t \to + \infty } \int_1^t {\frac{2}{x}} dx\\\int_1^\infty g (x)dx &= \mathop {\lim }\limits_{t \to + \infty } (2\ln |x|)_1^t\\\int_1^\infty g (x)dx &= \mathop {\lim }\limits_{t \to + \infty } (2\ln |t| - 2\ln |1|)\\\int_1^\infty g (x)dx &= + \infty \end{aligned}\)

Thus, the integral \(\int_1^\infty g (x)\) diverges.

Thus, by comparison theorem, the given function\(\int_1^\infty {\frac{{2 + {e^{ - \infty }}}}{x}} dx\)also diverges.