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Evaluate the integral$\int {\dfrac{{dt}}{t}} $ from $20 \to \infty $ if it converges.

Answer
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557.7k+ views
Hint:We have to find the integration with limits meaning definite integration is to be done here in the question. We know that integration of $\dfrac{1}{t}$ with respect to $t$ is the natural logarithm of $t$ i.e. $\int {\dfrac{1}{t}dt = \ln t + C} $; where $C$ is the integrating constant which will be removed when we use the limits.

Complete solution step by step:
Firstly we write down the given function which we are supposed to find the integration of i.e.
$I = \int {\dfrac{{dt}}{t}} $
We are given limits from $20$ to $\infty $ so if we apply the limits to the function we have
$I = \int\limits_{20}^\infty {\dfrac{1}{t}dt} $
Now to solve this integration we have to keep it mind that the formula for integrating reciprocal of $t$ with respect to is given by
$\int {\dfrac{1}{t}dt = \ln \left| t \right| + C} $ where $C$ is the integrating constant.
Now using this formula and applying the limits we have
$I = \int\limits_{20}^\infty {\dfrac{1}{t}dt = \left[ {\ln \left| t \right|} \right]} _{20}^\infty $
At this step we have to take the upper limit first then subtract the lower limit like this
$I = \left[ {\ln \infty - \ln 20} \right]$
Here we can see
We have a given condition in the question which says that if the function converges then only the integration is possible after putting the limits but we have $\ln \infty $ and we know that
$
\mathop {\lim }\limits_{x \to \infty } \ln x = \infty \\
\Rightarrow \mathop {\lim }\limits_{t \to \infty } \ln t = \infty \\
$
So the integral does not converge and we cannot find the integral with the given condition.

Note: Here in the question we applied the formula and the integration was easy to get but due to the given condition the integral comes out to be a non convergent one which means we cannot reach a point where we can say that the function will increase or decrease.