Question

Let $a_n(n \geq 1)$ be the value of $x$ for which $\int\limits_x^{2 x} e^{-t^n} d t(x>0)$ is maximum. If $L=\underset{n \rightarrow \infty}{\text{Lim}} \ln \left(a_n\right)$ then find the value of $e^{-L}$.

Answer 1

Let $f_n(x)=\int\limits_x^{2 x} e^{-t^2} d t$
$f_n^{\prime}(x)=2 \cdot e^{-(2 x)^n}-e^{-x^n}$
For maxima and minima, $f _{ n }^{\prime}( x )=0 \Rightarrow 2 e ^{-(2 x )^{ n }}= e ^{- x ^{ n }} \Rightarrow 2 \cdot e ^{-\left(2^{ n } x ^{ n }\right)}= e ^{- x ^{ n }}$ Taking log on both sides, we get
$\ln 2-2^{ n } x ^{ n }=- x ^{ n } \Rightarrow \ln 2= x ^{ n }\left(2^{ n }-1\right) \Rightarrow x ^{ n }=\frac{\ln 2}{2^{ n }-1} \Rightarrow x =\left(\frac{\ln 2}{2^{ n }-1}\right)^{\frac{1}{ n }}= a _{ n }$
Also, $f _{ n }^{\prime \prime}\left( x =\left(\frac{\ln 2}{2^{ n }-1}\right)^{\frac{1}{ n }}\right)<0 \Rightarrow f _{ n }( x )$ is maximum at $x =\left(\frac{\ln 2}{2^{ n }-1}\right)^{\frac{1}{ n }}$.
Now, $\ln a _{ n }=\frac{\ln \left(\frac{\ln 2}{2^{ n }-1}\right)}{ n }=\frac{\ln (\ln 2)-\ln \left(2^{ n }-1\right)}{ n }$
Hence $L =\underset{n \rightarrow \infty}{\text{Lim}} \ln \left( a _{ n }\right)=\underset{n \rightarrow \infty}{\text{Lim}} \frac{\ln (\ln 2)-\ln \left(2^{ n }-1\right)}{ n }=\underset{n \rightarrow \infty}{\text{Lim}}\left(\frac{\ln (\ln 2)}{ n }-\frac{\ln \left(2^{ n }\left(1-\frac{1}{2^{ n }}\right)\right)}{ n }\right) $
$L =\underset{n \rightarrow \infty}{\text{Lim}}\left(0-\frac{ n \cdot \ln 2+\ln \left(1-\frac{1}{2^{ n }}\right)}{ n }\right)=-\ln 2$
Hence, $e ^{- L }=2$