Question

Topic : Definite integral as a limit of a sum

Evaluate :-

$\boxed{\lim\limits_{x \to \infty} \left(\sum\limits_{n=1}^x \dfrac1 x \ln \left(\dfrac n x\right)\right)}$

Answer along with explanation

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\lim\limits_{x \to \infty} \left(\sum\limits_{n=1}^x \dfrac1 x \ln \left(\dfrac n x\right)\right)$

We know, using limit as a sum, it can be rewritten as

$\rm :\longmapsto\:\lim\limits_{x \to \infty}\sum\limits \: changes \: to \: sign \: of \: \: \displaystyle\int$

$\rm :\longmapsto\:a \: = \: \lim\limits_{x \to \infty} \: \dfrac{1}{x} \: = \: 0$

$\rm :\longmapsto\:b \: = \: \lim\limits_{x \to \infty} \: \dfrac{x}{x} \: = \: 1$

$\rm :\longmapsto\:\dfrac{1}{x} \: changes \: to \: dy$

$\rm :\longmapsto\:log\dfrac{n}{x} \: changes \: to \: logy$

So, above expression can be rewritten as

$\rm :\longmapsto\:\lim\limits_{x \to \infty} \left(\sum\limits_{n=1}^x \dfrac1 x \ln \left(\dfrac n x\right)\right) =  \: \displaystyle\int_0^1 \: logy \: dy$

Let evaluate first

$\rm :\longmapsto\:\displaystyle\int \: logy \: dy$

can be rewritten as

$\rm \:  =  \: \displaystyle\int \:1. logy \: dy$

Now, using integration by parts, we have

$\boxed{\tt{\displaystyle\int \: uvdx \: = u\displaystyle\int \: vdx \: - \: \displaystyle\int\bigg[\dfrac{d}{dx}u\displaystyle\int \: vdx \bigg] dx}}$

So, here,

$\rm :\longmapsto\:u = logy$

$\rm :\longmapsto\:v = 1$

$\rm \:  =  \: logy\displaystyle\int \: 1 \: dy - \displaystyle\int\bigg[\dfrac{d}{dy}logy\displaystyle\int \: 1dy \bigg]dy$

$\rm \:  =  \: y \: logy - \displaystyle\int \: \dfrac{1}{y} \times y \: dy$

$\rm \:  =  \: y \: logy - \displaystyle\int \: 1 \: dy$

$\rm \:  =  \: y \: logy - y$

Therefore,

$\rm :\longmapsto\: \: \displaystyle\int_0^1 \: logy \: dy$

$\rm \:  =  \: \bigg[y \: logy \: - \: y\bigg]_0^1$

$\rm \:  =  \: (log1 - 1) - (0 - 0)$

$\rm \:  =  \: (0 - 1)$

$\rm \:  =  \: - 1$

Hence,

$\rm :\longmapsto\:\boxed{ \: \: \tt{ \lim\limits_{x \to \infty} \left(\sum\limits_{n=1}^x \dfrac1 x \ln \left(\dfrac n x\right)\right) = - 1 \: \: }}$

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Additional Information

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$