Notes: `int_1^oo (x)/(2x^2-2c) Find a unique real number "c" which will allow the integral to converge. How would you do this problem?

Wednesday, July 27, 2016

$\displaystyle{\int_{{1}}^{\infty}}\frac{{{x}}}{{{2}{{x}}^{{2}}-{2}{c}}}{F}\in{d}{a}{u}{n}{i}{q}{u}{e}{r}{e}{a}{l}\nu{m}{b}{e}{r}\text{c}{w}{h}{i}{c}{h}{w}{i}{l}{l}{a}{l}{l}{o}{w}{t}{h}{e}\int{e}{g{{r}}}{a}{l}\to{c}{o}{n}{v}{e}{r}\ge.{H}{o}{w}{w}{o}{\underline{{d}}}{y}{o}{u}{d}{o}{t}{h}{i}{s}{p}{r}{o}{b}\le{m}?$

Hello!

Let's transform the expression under the integral sign:

$\displaystyle\frac{{x}}{{{2}{{x}}^{{2}}-{2}{c}}}-\frac{{c}}{{{x}+{1}}}=\frac{{{{x}}^{{2}}+{x}-{2}{c}{{x}}^{{2}}+{2}{{c}}^{{2}}}}{{{\left({2}{{x}}^{{2}}-{2}{c}\right)}{\left({x}+{1}\right)}}}=\frac{{{{x}}^{{2}}{\left({1}-{2}{c}\right)}+{x}+{2}{{c}}^{{2}}}}{{{\left({2}{{x}}^{{2}}-{2}{c}\right)}{\left({x}+{1}\right)}}}=$

$\displaystyle=\frac{{{{x}}^{{2}}{\left({1}-{2}{c}\right)}}}{{{\left({2}{{x}}^{{2}}-{2}{c}\right)}{\left({x}+{1}\right)}}}+\frac{{{x}+{2}{{c}}^{{2}}}}{{{\left({2}{{x}}^{{2}}-{2}{c}\right)}{\left({x}+{1}\right)}}}.$

The second fraction behaves as $\displaystyle\frac{{1}}{{{x}}^{{2}}}$ at infinity and therefore the integral of it converges at infinity, even absolutely. Consider the first fraction, it must converge also for the sum to converge.

For $\displaystyle{1}-{2}{c}\ne{0},$ i.e. for $\displaystyle{c}\ne\frac{{1}}{{2}},$ this fraction behaves like $\displaystyle\frac{{1}}{{x}}$ at infinity. More precisely, for $\displaystyle{1}-{2}{c}\gt{0},$ i.e. $\displaystyle{c}\lt\frac{{1}}{{2}},$

$\displaystyle\frac{{{{x}}^{{2}}{\left({1}-{2}{c}\right)}}}{{{\left({2}{{x}}^{{2}}-{c}\right)}{\left({x}+{1}\right)}}}\geq{\left({1}-{2}{c}\right)}\cdot\frac{{{{x}}^{{2}}}}{{{3}{{x}}^{{2}}\cdot{2}{x}}}=\frac{{{1}-{2}{c}}}{{6}}\cdot\frac{{1}}{{x}}$ for sufficiently large $\displaystyle{x}.$ So this integral diverges at infinity. The similar...

Hello!

Let's transform the expression under the integral sign:

So the only candidate is $\displaystyle{c}=\frac{{1}}{{2}},$ for this $\displaystyle{c}$ only the integral converges at infinity. But we have to check also the other possible critical points at $\displaystyle{\left[{1},\infty\right)}.$ They are the points where the denominator equals zero.

One of them is where $\displaystyle{x}+{1}={0},$ i.e. $\displaystyle{x}_{{1}}=-{1}.$ It is outside $\displaystyle{\left[{1},\infty\right)}.$ The next is where $\displaystyle{2}{{x}}^{{2}}-{2}{c}={2}{{x}}^{{2}}-{1}={0}.$ They are $\displaystyle{x}_{{2}}=\frac{{1}}{\sqrt{{{2}}}}$ and $\displaystyle{x}_{{3}}=-\frac{{1}}{\sqrt{{{2}}}},$ also outside the realm of integration. Good, no more critical points!