improper integrals problems

• Dec 29th 2010, 09:59 PM
Volga
improper integrals problems
I am an Exernal student (no in-person tuition or lectures), and I am practicing problems from my Calculus Study Guide, but the answers are not given. I'd appreciate if you could check my answers.

1. Determine whether the following integrals converge:

1.1 $\displaystyle \int^1_0\frac{cos x}{\sqrt{x}}dx$

Type 2 improper integral - problem spot: x=0

Since $\displaystyle \mid\cos x\mid\leq1$, compare the original integral to $\displaystyle \int^1_0\frac{1}{\sqrt{x}}dx$ which we know converges near 0 as the power of the denominator is less than 1.
Since $\displaystyle \int^1_0\frac{cos x}{\sqrt{x}}dx \leq \int^1_0\frac{1}{\sqrt{x}}dx$, which converges, the original integral also converges.

1.2 $\displaystyle \int^1_0\frac{\exp^{-x}}{\sqrt{x}}dx$

Type 2 improper integral - problem spot: x=0.

Choose a function $\displaystyle g(x)=\frac{1}{\sqrt {x}}$ which is similar to the original integrand $\displaystyle f(x)=\frac{\exp^{-x}}{\sqrt{x}}$ since
$\displaystyle lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\exp^{(-x)}=1$

$\displaystyle \int^1_0\frac{1}{\sqrt{x}}dx$ converges, and so does the original integral.

1.3 $\displaystyle \int^{\infty}_1\frac{\exp^{-x}}{\sqrt{x\ln(x+1)}}dx$

Type 1 improper integral - problem spot: infinity.

Chose a comparision function $\displaystyle g(x)=\frac{\exp(-x)}{\sqrtx}}$
$\displaystyle lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=\frac{1} {\sqrt{\ln{x+1}}}=0$ as the denominator goes to infinity

Since $\displaystyle \int^{\infty}_1\frac{\exp^{-x}}{\sqrt{x}}dx$ converges, so does the original integral, by one-way limit test.

1.4 $\displaystyle \int^1_0\frac{\exp^{-x}}{\sqrt{x\ln(x+1)}}dx$

Type 2 improper integral - problem spot: x=0.

Here I am a bit lost and would appreciate some hints. I know that the denominator is bounded (ln (x+1) ranges from 0 to ln2, and x ranges from 0 to 1. Does it help to chose a convergence test?

+I will post more in a follow up post.
• Dec 29th 2010, 10:12 PM
Volga
2. Determine whether the following integrals converge.

2.1 $\displaystyle \int^1_0\frac{cos x}{x^2}dx$

Improper integral 2nd type, problem at x=0.
Compare the integrand to $\displaystyle g(x)=\frac{1}{x^2} lim_{x\rightarrow0} \frac{\cosx}{x^2}:\frac{1}{x^2}=1$

Since the second integral diverges (p>1), so does the original one.

2.2 $\displaystyle \int^{\frac{\pi}{2}}_0\ln(sin x)dx$

Type 2 improper integral, problem spot at x=0.

Chose comparision function $\displaystyle g(x)=\ln(x)$

$\displaystyle lim_{x\rightarrow0}\frac{\ln(\sin(x)}{\ln(x)}=lim_ {x\rightarrow0}\frac{\cos(x)}{\sin(x)}:\frac{1}{x} =\frac{x}{\sin(x)}\cos(x)=1$ (using L'Hospital rule)

Therefore, the original function behaves similarly to g(x) near 0. Since g(x)=ln(x) go slowly to $\displaystyle -\infty$ as $\displaystyle x\rightarrow0$, it diverges, and so does the original integral.
• Dec 29th 2010, 10:56 PM
Volga
3. Discuss the convergence or otherwise of the following integrals.

3.1 $\displaystyle \int^1_0(\sin x)^pdx$

The behaviour of the integral depends on p.

If p>0, the integral is a proper integral which converges.

If P<0, there is a problem spot at x=0 (division by 0: $\displaystyle (\sin x)^p=\frac{1}{(\sin x)^{-p}}$)

Let p=-n and consider $\displaystyle \int^1_0\frac{1}{(\sin x)^n}dx$ where n is positive number.

From there, the only comparison function I could come up with is $\displaystyle \frac{x^n}{(\sin x)^n}$ which goes to 1 as x tends to 0, but when I compare $\displaystyle \int^1_0\frac{1}{(\sin x)^n}dx$ to that one, the first one is larger than the second, and it does not give me anything in terms of convergence.

need help here...

3.2 $\displaystyle \int^{\frac{\pi}{2}}_0\ln(\sin x)dx$

As x ranges from 0 to [tex]\frac{\pi}{2}, sin x goes from 0 to 1. Therefore, problem spot is at x=0.
As $\displaystyle x\rightarrow0, \sin x\rightarrow0$ and $\displaystyle \ln (\sin x)\rightarrow-\infty$. Therefore, integral diverges.

3.3 $\displaystyle \int^{\infty}_1\frac{\sqrt{x+1}-\sqrt{x}}{x}$

Split into two components,
$\displaystyle \int^{\infty}_1\frac{\sqrt{x+1}}{x}$ - $\displaystyle \int^{\infty}_1\frac{\sqrt{x}}{x}$ where the second component diverges after simplification:

$\displaystyle \int^{\infty}_1\frac{1}{\sqrt{x}}$ diverges as p<1.

Therefore, the whole original integral diverges.

3.4 $\displaystyle \int^{\pi}_0\frac{1}{x}\sin{\frac{1}{x}}dx$

Substitute $\displaystyle \frac{1}{x}=y$. This will change the limits of integration as follows:
$\displaystyle y=1/x\ x=0\ y=\infty \x=\pi\ y=\frac{1}{\pi}$
And $\displaystyle \frac{dy}{dx}=-\frac{1}{x^2}$

The integral becomes

$\displaystyle -\int_{\frac{1}{\pi}}^{\infty}\frac{1}{x}\sin{y}(-x^2)dy=\int_{\frac{1}{\pi}}^{\infty}\frac{\sin y}{y}dy$

then I can split this into two improper integrals (one from 1/pi to 1, another from 1 to infinity) and compare one of them to 1/y function, which diverges on either interval. Therefore, I can conclude that the original integral also diverges.
• Dec 29th 2010, 11:16 PM
FernandoRevilla
Quote:

Originally Posted by Volga
1.1 $\displaystyle \int^1_0\frac{cos x}{\sqrt{x}}dx$
Since $\displaystyle \mid\cos x\mid\leq1$, compare the original integral to $\displaystyle \int^1_0\frac{1}{\sqrt{x}}dx$ which we know converges near 0 as the power of the denominator is less than 1.
Since $\displaystyle \int^1_0\frac{cos x}{\sqrt{x}}dx \leq \int^1_0\frac{1}{\sqrt{x}}dx$, which converges, the original integral also converges.

Right, but you should say:

$\displaystyle 0\leq \dfrac{\cos x}{\sqrt{x}}\leq \dfrac{1}{\sqrt{x}}\Rightarrow \displaystyle\int_0^1\dfrac{\cos x}{\sqrt{x}}dx\;\textrm{convergent}$

Alternatively:

$\displaystyle \displaystyle\int_0^1\dfrac{\cos x}{\sqrt{x}}dx$

is absolutely convergent, as a consequence convergent.

Fernando Revilla
• Dec 29th 2010, 11:27 PM
FernandoRevilla
Quote:

Originally Posted by Volga
1.2 $\displaystyle \int^1_0\frac{\exp^{-x}}{\sqrt{x}}dx$
Choose a function $\displaystyle g(x)=\frac{1}{\sqrt {x}}$ which is similar to the original integrand $\displaystyle f(x)=\frac{\exp^{-x}}{\sqrt{x}}$ since
$\displaystyle lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\exp^{(-x)}=1$

$\displaystyle \int^1_0\frac{1}{\sqrt{x}}dx$ converges, and so does the original integral.

Right. Also:

$\displaystyle 0\leq\dfrac{e^{-x}}{\sqrt{x}}\leq \dfrac{1}{\sqrt{x}}\;\forall x\in (0,1]\Rightarrow \ldots$

Fernando Revilla
• Dec 29th 2010, 11:36 PM
FernandoRevilla
Quote:

Originally Posted by Volga
1.4 $\displaystyle \int^1_0\frac{\exp^{-x}}{\sqrt{x\ln(x+1)}}dx$ Here I am a bit lost and would appreciate some hints.

Use:

$\displaystyle \ln (x+1)\sim x\;(x\rightarrow 0)$

Fernando Revilla

P.S. Now, a rest. :)
• Dec 30th 2010, 12:29 AM
Volga
Thank you Fernando!

1.1 Got that! thanks for the feedback.

1.4 Then I can use $\displaystyle g(x)=\int^1_0\frac{\exp^{-x}}{x}dx$ which is similar to the original function f(x):

$\displaystyle lim_{x\rightarrow0}\frac{f(x)}{g(x)}=lim_{x\righta rrow0}(\frac{\exp^{-x}}{\sqrt{x\ln(x+1)}}:\frac{\exp(-x)}{x})=\frac{x}{x}=1$

Now, to evaluate g(x), I note that $\displaystyle \exp(-x)\rightarrow1$ as x approaches 0. Therefore, the 'dominant' component is 1/x which goes to infinity as x approaches 0. (can I put it this way, without showing any more additional comparison functions and limits?)

Overall, g(x) diverges, and so does the original integral.

Thank you for your help, and as I am not under time pressure, I'd appreciate if you look back here when you have free time.
• Dec 30th 2010, 01:04 AM
FernandoRevilla
Quote:

Originally Posted by Volga
1.4 Then I can use $\displaystyle g(x)=\int^1_0\frac{\exp^{-x}}{x}dx$ which is similar to the original function f(x):

It isn't necessary:

$\displaystyle \displaystyle\lim_{x \to 0}{\left(\dfrac{e^{-x}}{\sqrt{x\ln(x+1)}}:\dfrac{1}{x}\right)}=\displa ystyle\lim_{x \to 0}{\left(\dfrac{e^{-x}}{x}:\dfrac{1}{x}\right)}=1\neq 0$

and

$\displaystyle \displaystyle\int_0^1\dfrac{dx}{x}$

is divergent.

Fernando Revilla
• Dec 30th 2010, 05:00 PM
Volga
4 Determine whether the following integrals converge

4.1 $\displaystyle \int^{\infty}_2\frac{dx}{x(\ln x)^2}$

Type 1 improper integral, problem - behaviour at infinity.

I substitute y=ln(x) and dx=dy*x. Limits of integration become ln(2) and infinity.

New integral

$\displaystyle \int^{\infty}_{\ln 2}\frac{dy}{y^2}$ coverges (p=2>1) and therefore the original integral coverges, too.

4.2 $\displaystyle \int^{\infty}_{1}\frac{dx}{x^2\sin{\frac{1}{x}}}$

Type 1 improper integral, problem - behaviour at infinity.

Consider function $\displaystyle \frac{\frac{1}{x}}{\sin\frac{1}{x}}$ which $\displaystyle \rightarrow1$ as $\displaystyle x\rightarrow\infty$.

Since $\displaystyle 0\leq\frac{1}{x^2\sin{\frac{1}{x}}}\leq\frac{1}{x\ sin{\frac{1}{x}}}=\frac{\frac{1}{x}}{\sin\frac{1}{ x}}\leq1$, the original integral converges.
• Jan 2nd 2011, 09:37 PM
Volga
5. Question. Use L'Hospital Rule to determine the following limit

$\displaystyle lim_{x\rightarrow0}(\frac{1}{x}-\frac{1}{\ln(1+x)})$. (Hint: you will need to use a common denominator).

Hence determine whether the following integral converges:
$\displaystyle \int^1_0(\frac{1}{x}-\frac{1}{\ln(1+x)})dx$ converges.

By applying (repeatedly) L'Hospital I 'reduced' the expression under the limit sign to eventually -1/2.

Looking at the improper integral, I note that the problem spot is at x=0. Can I directly apply my conclusion about the limit of the integrand (when x->0) to the integral convergence? Ie since the function of the integrand is bounded (approaches -1/2 when x approaches 0), the integral itself is convergent?
• Jan 3rd 2011, 12:35 AM
FernandoRevilla
Quote:

Originally Posted by Volga
Can I directly apply my conclusion about the limit of the integrand (when x->0) to the integral convergence? Ie since the function of the integrand is bounded (approaches -1/2 when x approaches 0), the integral itself is convergent?

It is convergent because defining

$\displaystyle f(0)=-1/2$

you have

$\displaystyle \int_0^1f(x)\;dx$

being $\displaystyle f$ continuous in $\displaystyle [0,1]$.

Fernando Revilla
• Jan 3rd 2011, 02:42 AM
Volga
thank you Fernando for such a clear answer!