Question

If $I_n=\int \frac{ e ^{( n +1) x } dx }{\left(1+ e ^{ x }+\frac{ e ^{2 x }}{2 !}+\ldots \ldots+\frac{ e ^{ nx }}{ n !}\right)}=\lambda_{ n }\left( e ^{ x }-\ln \left( f _{ n }( x )\right)\right)+ C$ where $f_n(0)=1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots \ldots+\frac{1}{n !}$ and $C$ is constant of integration and $g(x)=\underset{n \rightarrow \infty}{\text{Lim}} \ln \left(f_n(x)\right)$, then find the number of real solutions of the equation $g(x)=4 x^2$.

Answer 1

Put $\quad e^x=t$
$I_n=\int \frac{t^n d t}{\left(1+t+\frac{t^2}{2 !}+\ldots \ldots+\frac{t^n}{n !}\right)}=n ! \int \frac{1+t+\frac{t^2}{2 !}+\ldots \ldots+\frac{t^n}{n !}-\left(1+t+\frac{t^2}{2 !}+\ldots \ldots+\frac{t^{n-1}}{(n-1) !}\right)}{\left(1+t+\frac{t^2}{2 !}+\ldots \ldots+\frac{t^n}{n !}\right)} d t$
$= n !\left( t -\int \frac{1+ t +\frac{ t ^2}{2 !}+\ldots \ldots+\frac{ t ^{ n -1}}{( n -1) !}}{1+ t +\frac{ t ^2}{2 !}+\ldots \ldots+\frac{ t ^{ n }}{ n !}} dt \right)$
Let $1+t+\frac{t^2}{2 !}+\ldots \ldots+\frac{t^n}{n !}=v ; d v=1+t+\frac{t^2}{2 !}+\ldots . .+\frac{t^{n-1}}{(n-1) !} d t$
$\therefore I_n=n !\left(t-\ln \left(1+t+\frac{t^2}{2 !}+\ldots \ldots+\frac{t^n}{n !}\right)\right)+c $
$=n !\left(e^x-\ln \left(1+e^x+\frac{e^{2 x}}{2 !}+\ldots \ldots+\frac{e^{n x}}{n !}\right)\right)+c $
$\therefore g ( x )=\underset { n \rightarrow \infty} {\text{Lim}}\ln \left(1+ e ^{ x }+\frac{ e ^{2 x }}{2 !}+\ldots \ldots+\frac{ e ^{ nx }}{ n !}\right)=\ln \left( e ^{ e ^{ x }}\right)= e ^{ x } $
$\therefore$ Number of solutions of $e ^{ x }= x ^2$ is 3