An interesting definite integral

Relevant wiki: Euler's number

$\int_{1}^{e}\ln (x)+\frac{1}{2}\left ( \ln (x)+\frac{1}{3}\left ( \ln (x)+\frac{1}{4}\left ( \ln (x)+\frac{1}{5}\left ( ... \right ) \right ) \right ) \right )dx$ = $\int_{1}^{e}\ln (x)+\frac{1}{2} \ln (x)+\frac{1}{6} \ln (x)+\frac{1}{24} \ln (x)...dx$ = $\int_{1}^{e}\ln (x) \left ( 1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}... \right )dx$ = $\int_{1}^{e}\ln (x)\times \left ( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}... \right )dx$

for the next step we must know that $\sum_{n=0}^{\infty } \frac{1}{n!}=e$

euler's number

so if $\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}...=e$ . then $\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}...=e-1$

$\int_{1}^{e}\ln (x)\times \left ( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}... \right )dx$ = $\int_{1}^{e}\ln (x) \left ( e-1 \right )dx$ = $\left ( e-1 \right )\int_{1}^{e}\ln (x)dx$ = $(e-1)\left [ x\ln (x)-x \right ]_{1}^{e}$ = $(e-1)\left [ (e\ln e-e)-(\ln 1-1) \right ]$ = $(e-1)\approx 1.7$

$\begin{aligned} & I = \displaystyle\int_{1}^{e} \left(\ln(x) +\dfrac{1}{2}(x) + \dfrac{1}{6}\ln(x) + \dfrac{1}{24}\ln(x) + \cdots \right)dx \\& I= \displaystyle\int_{1}^{e} \left(\ln(x) + \ln(x)^{\dfrac{1}{2}}+ \ln(x)^{\dfrac{1}{6}} + \ln(x)^{\dfrac{1}{24}}+\cdots \right)dx\qquad {\color{#3D99F6}\dfrac{1}{a}\ln(x)= \ln(x)^{\dfrac{1}{a}}} \\& I =\displaystyle\int_{1}^{e} \ln(x.x^{\frac{1}{2}}.x^{\frac{1}{6}}.x^{\frac{1}{24}}.\cdots)dx \phantom{cccc} {\color{#3D99F6} \ln(a)+\ln(b) = \ln(ab)} \\& I = \displaystyle\int_{1}^{e} \left(\ln(x)^{1+\dfrac{1}{2}+\dfrac{1}{6} + \dfrac{1}{24}+\cdots}\right) dx\\& I = \displaystyle\int_{1}^{e}\left(\ln(x)^{e-1}\right)\implies (e-1) \int_{1}^{e}\ln(x) dx \end{aligned}$

Now using Integration by parts and setting the upper and lower limit we find as $\begin{aligned} & I = (e-1)\displaystyle(e\ln(e) -(\ln(1)-1) = \boxed {e-1}\end{aligned}$

Define $f_n(x) = \ln x + \dfrac{1}{2} \left ( \ln x + \dfrac{1}{3} \left ( \ln x + \cdots + \dfrac{1}{n} \ln x \right ) \cdots \right ).$ Notice that

$\begin{aligned} f_n(x) &= \ln x + \dfrac{1}{2} \left ( \ln x + \dfrac{1}{3} \left ( \ln x + \cdots + \dfrac{1}{n} \ln x \right ) \cdots \right ) \\ &= \ln x + \dfrac{1}{2} \left ( \ln x + \dfrac{1}{3} \left ( \ln x + \cdots + \dfrac{1}{n - 1} \left ( \ln x + \dfrac{1}{n} \ln x \right ) \right ) \cdots \right ) \\ &= \ln x + \dfrac{1}{2} \left ( \ln x + \dfrac{1}{3} \left ( \ln x + \cdots + \dfrac{1}{n - 1} \ln x \right ) \cdots \right ) + \dfrac{1}{n!} \ln x \\ &= f_{n - 1}(x) + \dfrac{1}{n!} \ln x. \end{aligned}$

Since $f_1(x) = \ln x,$ a quick induction gives us $f_n(x) = \left ( 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots + \dfrac{1}{n!} \right ) \ln x.$ Thus,

$\begin{aligned} \int_1^e \left ( \ln x + \dfrac{1}{2} \left ( \ln x + \dfrac{1}{3} ( \ln x + \cdots ) \cdots \right ) \right ) \, dx &= \int_1^e \lim_{n \rightarrow \infty} f_n(x) \, dx \\ &= \int_1^e \lim_{n \rightarrow \infty} \left [ \left ( 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots + \dfrac{1}{n!} \right ) \ln x \right ] \, dx \\ &= \int_1^e (e - 1) \ln x \, dx \\ &= (e - 1) [x \ln x - x]_1^e \\ &= (e - 1)((e \ln e - e) - (1 \ln 1 - 1)) \\ &= \boxed{e - 1}. \end{aligned}$

If you're anything like me and have no idea what you're doing, obtain bounds on the integral and use those to conveniently eliminate all choices except the correct one. $\ln x < \ln x + \frac{1}{2} \left(\ln x + \frac{1}{3} \left( \ln x + \cdots\right)\right) < \ln x + \frac{1}{2} \left(\ln x + \frac{1}{2} \left( \ln x + \cdots\right)\right)$ $\Rightarrow \ln x < \ln x + \frac{1}{2} \left(\ln x + \frac{1}{3} \left( \ln x + \cdots\right)\right) < 2 \ln x$

Integrating $\ln x$ from 1 to $e$ yields $1$ and integrating $2 \ln x$ yields $2$ . Looking at the choices, we have $e = 2.71\cdots$ , $e + 1 = 3.71 \cdots$ , $1/e < 1$ . Of course, the only possible answer to choose from, then, is $\boxed{e - 1}$ .