### $I = \int\limits_0^\infty {\frac{{{e^{ - nx}} - {e^{ - x}}}}{x}dx} = ?$$n \in \mathbb{N} Let, {e^{ - x}} = t \Rightarrow - {e^{ - x}}dx = dt \Rightarrow - tdx = dt I = \int\limits_1^0 {\frac{{{t^n} - t}}{{ - \ln t}}\frac{{dt}}{{ - t}}} = - \int\limits_0^1 {\frac{{{t^{n - 1}} - 1}}{{\ln t}}dt} But, I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } - 1}}{{\ln x}}dx = \ln (\lambda + 1)} (Proof can be seen in this problem) So, I = - \ln (n - 1 + 1) = - \ln n ### Popular posts from this blog ### f(x)=x^6+2x^4+x^3+2x+3$$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$
Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R$. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$