Improper integrals (1 Viewer)

leehuan · May 6, 2016

$Question: Does $\int_{-\infty}^{\infty}{\frac{x}{1+x^2}dx}$ converge? Explain.$

The answer is obviously that it does not, but I want to check if my proof is flawed.

$$Suppose $f(x)=\frac{x}{1+x^2}$ and $g(x)=\frac{1}{x}$. Now, \\ $\lim_{x\rightarrow \infty}{\frac{f(x)}{g(x)}}=\dots=1$

$$And $\int_{-\infty}^{\infty}{\frac{x}{1+x^2}dx}=\int_{-\infty}^{1}{\frac{x}{1+x^2}dx}+\int_{1}^{\infty}{ \frac{x}{1+x^2} dx}$

$$Since $\lim_{x\rightarrow \infty}{\frac{f(x)}{g(x)}}$ is defined for a specific $0<a<\infty \quad (a=1)$ and by the $p$-test, $\int_1^{\infty}{g(x) dx}$ diverges,\\ by the limit form of the comparison test $\int_{1}^{\infty}{\frac{x}{1+x^2}dx}$ also diverges, hence so must $\int_{-\infty}^{\infty}{\frac{x}{1+x^2}dx}$

I have a feeling I may have needed to consider the negative infinity portion of this integral?

InteGrand · May 6, 2016

leehuan said:
$Question: Does $\int_{-\infty}^{\infty}{\frac{x}{1+x^2}dx}$ converge? Explain.$

The answer is obviously that it does not, but I want to check if my proof is flawed.

$$Suppose $f(x)=\frac{x}{1+x^2}$ and $g(x)=\frac{1}{x}$. Now, \\ $\lim_{x\rightarrow \infty}{\frac{f(x)}{g(x)}}=\dots=1$

$$And $\int_{-\infty}^{\infty}{\frac{x}{1+x^2}dx}=\int_{-\infty}^{1}{\frac{x}{1+x^2}dx}+\int_{1}^{\infty}{ \frac{x}{1+x^2} dx}$

$$Since $\lim_{x\rightarrow \infty}{\frac{f(x)}{g(x)}}$ is defined for a specific $0<a<\infty \quad (a=1)$ and by the $p$-test, $\int_1^{\infty}{g(x) dx}$ diverges,\\ by the limit form of the comparison test $\int_{1}^{\infty}{\frac{x}{1+x^2}dx}$ also diverges, hence so must $\int_{-\infty}^{\infty}{\frac{x}{1+x^2}dx}$

I have a feeling I may have needed to consider the negative infinity portion of this integral?

The Cauchy principal value would be 0. Otherwise it's an undefined expression -inf + inf.

Don't need to use any comparison tests, we can find an antiderivative using inspection or a u-substitution.

Paradoxica · May 6, 2016

leehuan said:
$Question: Does $\int_{-\infty}^{\infty}{\frac{x}{1+x^2}dx}$ converge? Explain.$

The answer is obviously that it does not, but I want to check if my proof is flawed.

$$Suppose $f(x)=\frac{x}{1+x^2}$ and $g(x)=\frac{1}{x}$. Now, \\ $\lim_{x\rightarrow \infty}{\frac{f(x)}{g(x)}}=\dots=1$

$$And $\int_{-\infty}^{\infty}{\frac{x}{1+x^2}dx}=\int_{-\infty}^{1}{\frac{x}{1+x^2}dx}+\int_{1}^{\infty}{ \frac{x}{1+x^2} dx}$

$$Since $\lim_{x\rightarrow \infty}{\frac{f(x)}{g(x)}}$ is defined for a specific $0<a<\infty \quad (a=1)$ and by the $p$-test, $\int_1^{\infty}{g(x) dx}$ diverges,\\ by the limit form of the comparison test $\int_{1}^{\infty}{\frac{x}{1+x^2}dx}$ also diverges, hence so must $\int_{-\infty}^{\infty}{\frac{x}{1+x^2}dx}$

I have a feeling I may have needed to consider the negative infinity portion of this integral?

Does it make sense to ask convergence of an integral whose CPV is 0?

In any case, you can reduce the two cases to one by using the parity of the function to your advantage.

leehuan · May 6, 2016

They didn't teach us about the Cauchy principal value. I only found that on WolframAlpha.

Edit: Though they did make us evaluate this
$\lim_{R\rightarrow \infty}{\int_{-R}^{R}{\frac{x}{1+x^2}dx}}$

Paradoxica · May 6, 2016

leehuan said:
They didn't teach us about the Cauchy principal value. I only found that on WolframAlpha.

Edit: Though they did make us evaluate this
$\lim_{R\rightarrow \infty}{\int_{-R}^{R}{\frac{x}{1+x^2}dx}}$

yeah, that's just the CPV.

leehuan · May 6, 2016

Paradoxica said:
yeah, that's just the CPV.

I suppose the question was just meant to distinguish between the two. But like I said, our course makes no reference to CPV.

leehuan · May 6, 2016

Bit of guidance needed here.

$Find all real numbers $p$ such that $\int_2^\infty{\frac{dx}{x {\left(\ln{x}\right)}^{p}}}$ converges.$

Drongoski · May 6, 2016

Typo? 'a' is 'p'?

leehuan · May 6, 2016

Drongoski said:
Typo? 'a' is 'p'?

My bad. Question still holds out of my fatigue though please?

InteGrand · May 6, 2016

leehuan said:
Bit of guidance needed here.

$Find all real numbers $p$ such that $\int_2^\infty{\frac{dx}{x {\left(\ln{x}\right)}^{p}}}$ converges.$

It ends up just being the p-test when you do a change of variables (substitute u = ln(x)).

So it converges iff p > 1.

leehuan · May 9, 2016

$$Use the transformation $t=x^{-1}$ to show that $\Gamma(s)=\int_0^\infty{e^{-t} t^{s-1} dt}$ is convergent for $s>0$
(I know this is the Gamma function)

I had an idea but then I got lost. My last step was rewriting the integral as

$\int_0^\infty{\frac{e^{-\frac{1}{x}}}{x^{s+1}}dx}$

InteGrand · May 9, 2016

leehuan said:
$$Use the transformation $t=x^{-1}$ to show that $\Gamma(s)=\int_0^\infty{e^{-t} t^{s-1} dt}$ is convergent for $s>0$
(I know this is the Gamma function)

I had an idea but then I got lost. My last step was rewriting the integral as

$\int_0^\infty{\frac{e^{\frac{1}{x}}}{x^{s+1}}dx}$

The -t in the exponential should become -1/x , rather than 1/x.

Drongoski · May 9, 2016

In transformed integral: $e^{-\frac {1}{x}}$ ??

ps
InteGrand beaten me to it.

seanieg89 · May 9, 2016

So you have the integrand $g(x)=x^{-1-s}e^{-x^{-1}}$ and we want to show it is integrable on the non-negative reals.

The behaviour near zero is not a problem at all, since $g(x)\rightarrow 0$ as $x\rightarrow 0$ (essentially because exponentials dominate polynomials, there are many ways you could justify this). So it suffices to check that g be integrable on $[1,\infty)$ .

This follows by comparison to $x^{-1-s}$ . (In fact we can use this argument to show that the integral is convergent if and ONLY if $s>0$ ).

leehuan · May 9, 2016

InteGrand said:
The -t in the exponential should become -1/x , rather than 1/x.

Mistake in typing, my bad

leehuan · May 30, 2016

Having already shown $\int_0^x{2te^{t^2}dt}=e^{x^2}-1$

$$Show that $\int_0^x{\left(t^2+t\right)e^{t^2}dt}\to \infty$ as $x \to \infty$

The method the answers provided is not immediately obvious to me and I don't understand the lateral thinking required. Can someone please explain the intuition required to see this method or just provide an easier pathway?

$$Answer given: For $t>1, \quad 2t \le t^2+t$

$$Hence for $x>1\\ \begin{align*} \int_0^x{\left(t^2+t\right)e^{t^2}dt}&=\int_0^1{\left(t^2+t\right)e^{t^2}dt}+\int_1^x{ \left( t^2+t \right) e^{t^2}dt}\\ & \ge \int_0^1{\left(t^2+t\right)e^{t^2}dt} + \int_1^x{2te^{t^2}dt} \\ &= \int_0^1{\left(t^2-t\right)e^{t^2}dt} + \int_0^x{2te^{t^2}dt}\\ &= \int_0^1{\left(t^2-t\right)e^{t^2}dt} + e^{x^2}-1 \to \infty \end{align*}$

InteGrand · May 30, 2016

leehuan said:
Having already shown $\int_0^x{2te^{t^2}dt}=e^{x^2}-1$

$$Show that $\int_0^x{\left(t^2+t\right)e^{t^2}dt}\to \infty$ as $x \to \infty$

The method the answers provided is not immediately obvious to me and I don't understand the lateral thinking required. Can someone please explain the intuition required to see this method or just provide an easier pathway?

$$Answer given: For $t>1, \quad 2t \le t^2+t$

$$Hence for $x>1\\ \begin{align*} \int_0^x{\left(t^2+t\right)e^{t^2}dt}&=\int_0^1{\left(t^2+t\right)e^{t^2}dt}+\int_1^x{ \left( t^2+t \right) e^{t^2}dt}\\ & \ge \int_0^1{\left(t^2+t\right)e^{t^2}dt} + \int_1^x{2te^{t^2}dt} \\ &= \int_0^1{\left(t^2-t\right)e^{t^2}dt} + \int_0^x{2te^{t^2}dt}\\ &= \int_0^1{\left(t^2-t\right)e^{t^2}dt} + e^{x^2}-1 \to \infty \end{align*}$

Yeah it's essentially a comparison test.

For all t large enough, the integrand in the second integral is greater than the first (since t^2 + t > 2t for all t large enough, say all t > 1).

Since the former integral taken from 1 to oo diverges to +oo (from previous part of Q.), by the comparison test, so does the latter one when taken from 1 to oo.

And remembering the finite lower limit here is irrelevant since the integrand is continuous, we're done.

Improper integrals (1 Viewer)

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