Matematické Fórum

stuart clark · 17. 10. 2019 16:23

Finding Convergency of Integral $\int^{\infty}_{0}\frac{(\ln x)^4}{e^{x^2}}dx$

laszky · 17. 10. 2019 21:45 — Editoval laszky (17. 10. 2019 22:06)

↑ krakonoš:

Hi.

You probably wanted to use a substitution $x=\sqrt{k}$ , nevertheless
$\ln x < x,\ \mbox{but}\ (\ln x)^4 \not < x^4\ \; \mbox{for all}\ x\in\mathbb{R}^{+}$ .

krakonoš

↑ laszky:
Ahoj.
Diky.
Myslela jsem puvodne na okoli nekonecna.Jeste se to ale bude muset doresit na okoli nuly.Na intervalu 0;K bude asi stacit zkoumat konvergenci majoritniho integralu (ln x) na ctvrtou.Po zvoleni substituce za ln x by to melo konvergovat.- per partes ctvrta mocnina s exponencielouVlastne by to melo i stacit pro cely zadany interval.

krakonoš

↑ stuart clark:
Hi

I= $\int_{K}^{\infty }\frac{(lnx)^{4}}{e^{x^{2}}}dx<$ $\int_{K}^{\infty }\frac{(x)^{4}}{e^{x^{2}}}dx$ , where $(lnx)^{4}<x^{4}$ for all x>K
You can use the substitution $y=x^{2}$ , you will get gamma function (konvergent majorant).

$\int_{0}^{K}\frac{(lnx)^{4}}{e^{x^{2}}}dx$ $<\int_{0}^{K}(lnx)^{4}dx$
After the substitution lnx=z you can use per partes.
Your integral is konvergent.