# Tag Info

3

According to Theorem 5.13 in Rudin's "Principles of mathematical analysis" (slightly restated) If $f$ and $h$ are real and differentiable in $(0,1)$ and if $h'(x)\neq 0$ in a neighborhood of $1$, and $h(x)\to +\infty$ as $x\to 1^-$, and in addition $f'(x)/h'(x)\to A$ as $x\to 1^-$, then $f(x)/h(x)\to A$ as $x\to 1^-$. Applying this to $$f(x)=\... 0 One way to express this integral as a function of the Elliptic Integral K, is as follows. Do the change of variables x^2 = \frac{x_0+^2}{1-t^2}. This converts the integral into :$$ \frac{1}{\sqrt{x_+^2-x_-^2}}\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1+\frac{x_-^2}{x_+^2-x_-^2}t^2)}} $$Which can be written formally as \frac{1}{\sqrt{x_+^2-x_-^2}}K(-\frac{x_-^2}{... 0 If$$f_h=\frac{1}{\sqrt{2 \pi \sigma ^2}}e^{-\frac{x^2}{2 \sigma ^2}}\qquad \text{and}\qquad f_\rho=\frac{f_h\left(\sqrt{\frac{x}{\rho }}\right)}{\sqrt{\rho x}}$$then$$f_\rho=\frac{e^{-\frac{x}{2 \rho \sigma ^2}}}{\sqrt{2 \pi \rho \sigma ^2} \sqrt{ x}}$$Let$$t^2=\frac{x}{2 \rho \sigma ^2}\implies x=2 \rho \sigma ^2 t^2\implies dx=4 \rho \sigma ^...

0


3


0


2

By using the comparison test, we know that if we integrate a rational function $\int_a^{\infty} p(x)/q(x)dx$ such that $q(x)\neq 0$ on the interval, then we converge if and only if $\deg(p)-\deg(q)<-1$. However, actually combining and simplifying the rational functions in this problem is tedious algebra, so one can ask if there is a better way. In fact, ...

0

I have finally managed to answer the question :) It is trivial to see that the integral is divergent for $r = 0\$, so let $r \neq 0$. If $r \neq 0$, note that we can always choose $a$ large enough such that $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x = \int^a_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x + \int^{\... 1 Try to bring the logarithms together! :)$$\lim_{a\to \infty} \biggl(r\ln\frac{a+1}{2}+\frac 3 4\ln|\frac{2+r}{2a^2+r}|\biggl)$$You maybe should take r < -2 and r > -2 because of the absolute value, but it gives the same result so I'll go with normal brackets.$$\lim_{a\to \infty} (\ln(\frac{a+1}{2})^r+ \ln(\frac{2+r}{2a^2+r})^\frac 3 4 )\...

0


1


3

$\newcommand{\S}{\operatorname{sinc}}$ A 'brute force' solution to a beautiful problem. I don't claim this answer is as insightful as the others and the question is somewhat old, but I feel it is relevant and unique enough to merit posting. As usual, we will let $\S(x) = \sin(x)/x$ with $\S(0)=1$. The big idea: use angle-addition, Taylor series, and ...

0


Top 50 recent answers are included