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Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite Background Consider the following divergent integral: $$ int_-infty^infty e^-e^x e^ax x^b dx $$ Where $a$ and $b$ are positive integers such that $b > a$. Multipling $e^-x cdot e^x$ inside the integral. $$ int_-infty^infty e^-e^x e^x e^(a-1)x x^b dx $$ Using integration by parts and noticing $(e^-e^x)' = -e^-e^xe^x $: $$ implies - int_-infty^infty (e^-e^x)' e^(a-1)x x^b dx = - int_-infty^infty e^-e^x (e^(a-1)x x^b)' dx $$ Again multiplying $e^-x cdot e^x$ inside the integral and repeating the process: $$ = - int_-infty^infty e^-e^x e^x cdot e^-x(e^(a-1)x x^b)' dx $$ $$ = int_-infty^infty (e^-e^x)'cdot e^-x(e^(a-1)x x^b)' dx $$ $$ = int_-infty^infty (e^-e^x)cdot (e^-x(e^(a-1)x x^b)')' dx $$ This trick will only work $a$ times. The leading order term in $x$ will be something like: $ a!int_-infty^infty e^-e^x x^b dx $ My idea was to take: $$ lim_n to infty fracint_-n^n e^-e^x e^ax x

Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.8 Suppose $f$ is holomorphic in region $G$, and $f(G) subseteq =1 $. Prove $f$ is constant. I will now attempt to elaborate the following proof at a Winter 2017 course in Oregon State University. What errors if any are there? Or are there more elegant ways to approach this? I have a feeling this can be answered with Ch2 only, i.e. Cauchy-Riemann or differentiation/holomorphic properties instead of having to use Möbius transformations. OSU Pf: Let $g(z)=frac1+z1-z$, and define $h(z)=g(f(z)), z in G setminus z : f(z) = 1$. Then $h$ is holomorphic on its domain, and $h$ is imaginary valued by Exer 3.7. By a variation of Exer 2.19, $h$ is constant. QED My Pf: $because f(G) subseteq C[0,1]$, let's consider the Möbius transformation in the preceding Exer 3.7 $g: mathbb C setminus z = 1 to mathbb C$ s