A couple of Improper Intergrals i have problems with

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In summary: No, those aren't the right limits of integration. Those are values of x, but u = x2.You can do either of the following:1. Change the x limits to the appropriate values of u.2. Undo the u substitution.Ok, so let's try this again. You said you did a u-substitution of u = x^2 and got the integral \frac{1}{2}\int\frac{1}{sen(u)} \, du = \frac{1}{2}ln|csc(u) + cot(u)| + C. Is that right?If so, then the original integral becomes \int^{\frac{\pi}{2}}_{0}\frac{1}{2
  • #1
SclayP
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INTEGRALS​

1.[itex]\int^{\frac{\Pi}{2}}_{0} \frac{x}{sin(x^2)} \, dx[/itex]

2.[itex]\int^{\infty}_{-\infty} \frac{1}{t} \, dt[/itex]

3.[itex]\int^{2}_{1} \frac{1}{xln(x)^4} \, dx[/itex]

RESOLVING THE INTEGRALS​

1. I firts resolved the definte Integral and then applied the
[tex]\lim_{b\rightarrow +0} -\frac{1}{2}ln(cotg(\frac{x^2}{2})|^{\frac{\pi}{2}}_{b} = \infty[/tex]

So, i come up to the resault that the integral does not converge. But i don't know if this procedure is ok, and if there is some way to do it by comparison.

2.So, this one i couldn't finish, you see, i come up to the next limit:

[tex]\lim_{b\rightarrow +\infty} -ln |-b|+ ln |b| = ?[/tex]

So, i don't know how to resolve the limit or how to analyse the integral by comparison.

3.Well, with this one i just don't know what to do...

Thanks, and sorry for my english.
 
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  • #2
SclayP said:
INTEGRALS​

1.[itex]\int^{\frac{\Pi}{2}}_{0} \frac{x}{sin(x^2)} \, dx[/itex]

2.[itex]\int^{\infty}_{-\infty} \frac{1}{t} \, dt[/itex]

3.[itex]\int^{2}_{1} \frac{1}{xln(x)^4} \, dx[/itex]

RESOLVING THE INTEGRALS​

1. I firts resolved the definte Integral and then applied the
[tex]\lim_{b\rightarrow +0} -\frac{1}{2}ln(cotg(\frac{x^2}{2})|^{\frac{\pi}{2}}_{b} = \infty[/tex]
Your antiderivative is wrong here. Show us how you got from the indefinite integral to your antiderivative. We can worry about the definite integral when you get the integration right.
SclayP said:
So, i come up to the resault that the integral does not converge. But i don't know if this procedure is ok, and if there is some way to do it by comparison.

2.So, this one i couldn't finish, you see, i come up to the next limit:

[tex]\lim_{b\rightarrow +\infty} -ln |-b|+ ln |b| = ?[/tex]
SclayP said:
So, i don't know how to resolve the limit or how to analyse the integral by comparison.

3.Well, with this one i just don't know what to do...

Thanks, and sorry for my english.
 
  • #3
Mark44 said:
Your antiderivative is wrong here. Show us how you got from the indefinite integral to your antiderivative. We can worry about the definite integral when you get the integration right.

To be honest i started de Integral by doing the subtitution
[tex]u = x^2[/tex][tex]du = 2x[/tex]
So,

[itex]\int \frac{x}{sen(x^2)} \, dx = \frac{1}{2}\int^{\frac{\pi^2}{4}_{1}\frac{1}{sen(u)} \, dx[/itex]

Then I looked up in the web "Wolfram|Alpha" how to proceed but they just jump from that integral i wrote, to the one you say it`s wrong. Later with the help of a PF MENTOR i came up to this Integral:

[itex]-\frac{1}{2}\int^{1}_{-0.78}\frac{1}{1-v^2} \, dv[/itex]

v = cos(u)
dv = -sen(u)
 
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  • #4
1/sin(u) can also be written csc(u). Probably best to just look up the integral of that. Or ask Wolfram to get the indefinite integral. Wolfram will also tell you the original integral diverges, but don't use that part. You should figure out why it diverges yourself.
 
  • #5
Dick said:
1/sin(u) can also be written csc(u). Probably best to just look up the integral of that. Or ask Wolfram to get the indefinite integral. Wolfram will also tell you the original integral diverges, but don't use that part. You should figure out why it diverges yourself.

Yes, of course. Well that was what actually did, i mean i subtituted u=x2, du=2xdx to get [itex]\frac{1}{2}\int\frac{1}{sen(u)} \, du[/itex] which is the integral of csc(u) and looked it up on Wolfram and it gives me the funcion [itex]f(x) = ln(sen(\frac{u}{2}))-ln(cos(\frac{u}{2})) + C[/itex].

So now i apply
[tex]\lim_{b\rightarrow 0} {ln(sen(\frac{x^2}{2}))-ln(cos(\frac{x^2}{2})) + C}|^{\frac{\pi}{2}}_{b}[/tex]

Then
[tex]\lim_{b\rightarrow 0} {ln(0.94) - ln(sen(\frac{b^2}{2})) - ln(0.34) + ln(cos(\frac{b^2}{2}))}[/tex]

And so that is equal to [itex]\infty[/itex].

That mean that does not converge but i wish there was another way to prove it does not converge like by comparison ( i don't know if its called like that in english) and not to do all the integral.

Sorry for my english and if there is some mistakes its reaaly hard to write math equation in here.

Thanks again
 
  • #6
SclayP said:
Yes, of course. Well that was what actually did, i mean i subtituted u=x2, du=2xdx to get [itex]\frac{1}{2}\int\frac{1}{sen(u)} \, du[/itex] which is the integral of csc(u) and looked it up on Wolfram and it gives me the funcion [itex]f(x) = ln(sen(\frac{u}{2}))-ln(cos(\frac{u}{2})) + C[/itex].
I don't know where you're getting what you show above. This is how the integral usually appears.
$$ 1/2 \int csc(u) du = -(1/2) ln|csc(u) + cot(u)| + C$$

I think you might be making errors in your work.
SclayP said:
So now i apply
[tex]\lim_{b\rightarrow 0} {ln(sen(\frac{x^2}{2}))-ln(cos(\frac{x^2}{2})) + C}|^{\frac{\pi}{2}}_{b}[/tex]

Then
[tex]\lim_{b\rightarrow 0} {ln(0.94) - ln(sen(\frac{b^2}{2})) - ln(0.34) + ln(cos(\frac{b^2}{2}))}[/tex]

And so that is equal to [itex]\infty[/itex].

That mean that does not converge but i wish there was another way to prove it does not converge like by comparison ( i don't know if its called like that in english) and not to do all the integral.

Sorry for my english and if there is some mistakes its reaaly hard to write math equation in here.

Thanks again
 
  • #7
Mark44 said:
I don't know where you're getting what you show above. This is how the integral usually appears.
$$ 1/2 \int csc(u) du = -(1/2) ln|csc(u) + cot(u)| + C$$

I think you might be making errors in your work.

YES, your absolutly right..Im getting it from Wolfram the thing is that what I'm writing is the equivalent for resticted u values. I'm sorry. So...

[itex]\int^{\frac{\pi}{2}}_{0}csc(u) \, du = -\frac{1}{2}ln(csc(u) + cot(u)) + C |^{\frac{\pi}{2}}_{0}[/itex]

Then,
[tex]\lim_{b\rightarrow 0} {-\frac{1}{2}ln|csc(u) + cot(u)| + C |^{\frac{\pi}{2}}_{b}}[/tex]

[tex]\lim_{b \to 0} -\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|csc(b^2) + cot(b^2)|[/tex]

So that is equal to

[itex]-\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|\infty| = \infty[/itex]

Thanks again...
 
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  • #8
SclayP said:
YES, your absolutly right..Im getting it from Wolfram the thing is that what I'm writing is the equivalent for resticted u values. I'm sorry. So...

[itex]\int^{\frac{\pi}{2}}_{0}csc(u) \, du = -\frac{1}{2}ln(csc(u) + cot(u)) + C |^{\frac{\pi}{2}}_{0}[/itex]
No, those aren't the right limits of integration. Those are values of x, but u = x2.

You can do either of the following:
1. Change the x limits to the appropriate values of u.
2. Undo the u substitution.
SclayP said:
Then,
[tex]\lim_{b\rightarrow 0} {-\frac{1}{2}ln|csc(u) + cot(u)| + C |^{\frac{\pi}{2}}_{b}}[/tex]

[tex]\lim_{b\rightarrow 0} -\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|csc(b^2) + cot(b^2)|[/tex]

So that is equal to

[itex]-\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|\infty| = \infty[/itex]

Thanks again...
 
  • #9
Mark44 said:
No, those aren't the right limits of integration. Those are values of x, but u = x2.

You can do either of the following:
1. Change the x limits to the appropriate values of u.
2. Undo the u substitution.

I actually wrote the wrong limits of integration but the math i did it right. i just wrote it wrongly,but yes it would be
[itex]\int^{\frac{\pi}{2}}_{0}csc(x^2) \, du = -\frac{1}{2}ln(csc(x^2) + cot(x^2)) + C |^{\frac{\pi}{2}}_{0}[/itex]

Then
[tex]\lim_{b\rightarrow 0} {-\frac{1}{2}ln|csc(x^2) + cot(x^2)| + C |^{\frac{\pi}{2}}_{b}}[/tex]
[tex]\lim_{b \to 0} -\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|csc(b^2) + cot(b^2)|[/tex]
 
  • #10
And it looks like both terms in the 2nd ln expression approach infinity.
 
  • #11
Mark44 said:
And it looks like both terms in the 2nd ln expression approach infinity.

So the integral does not converge after all...
 
  • #12
Mark44 said:
I don't know where you're getting what you show above. This is how the integral usually appears.
$$ 1/2 \int csc(u) du = -(1/2) ln|csc(u) + cot(u)| + C$$

I think you might be making errors in your work.

Wolfram Alpha does give log(sin(u/2))+log(cos(u/2)) as the antiderivative instead of the more usual -log(csc(u)+cot(u)). There's nothing wrong with SclaP's work. I suggested looking it up rather than deriving it. Since that's what I would do and the main point is to evaluate the limit. Both give the same result.
 

FAQ: A couple of Improper Intergrals i have problems with

What are improper integrals?

Improper integrals are definite integrals where either the upper or lower limit of integration is infinite, or the function being integrated becomes unbounded at one or more points within the interval of integration.

Why are improper integrals important?

Improper integrals allow us to calculate the areas under curves that would not be possible with regular definite integrals. They are also crucial in many areas of science and engineering, such as in calculating probabilities and solving differential equations.

How do you determine the convergence of an improper integral?

To determine the convergence of an improper integral, you must evaluate the limit of the integral as one or both of the limits of integration approaches infinity. If the limit exists and is finite, then the integral converges. If the limit does not exist or is infinite, then the integral diverges.

What is the difference between a type 1 and type 2 improper integral?

Type 1 improper integrals have one or both limits of integration that are infinite, while type 2 improper integrals have a function that becomes unbounded at one or more points within the interval of integration. Type 1 improper integrals are evaluated by taking the limit of the integral as the limits of integration approach infinity, while type 2 improper integrals are evaluated by breaking the integral into smaller intervals where the function is bounded and then taking the limit as the upper limit of integration approaches the point where the function becomes unbounded.

How do you solve improper integrals?

To solve improper integrals, you must first determine if the integral converges or diverges. If it converges, you can evaluate the limit of the integral to find the exact value. If it diverges, you can use different techniques such as finding an appropriate substitution or using comparison tests to determine the behavior of the integral. In some cases, improper integrals may require more advanced techniques such as partial fractions or trigonometric substitutions to evaluate.

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