A couple of Improper Intergrals i have problems with

[SclayP]

INTEGRALS​

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

2.\int^{\infty}_{-\infty} \frac{1}{t} \, dt

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

RESOLVING THE INTEGRALS​

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

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:

\lim_{b\rightarrow +\infty} -ln |-b|+ ln |b| = ?

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.

 

[Mark44]

INTEGRALS​

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

2.\int^{\infty}_{-\infty} \frac{1}{t} \, dt

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

RESOLVING THE INTEGRALS​

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

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.

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:

\lim_{b\rightarrow +\infty} -ln |-b|+ ln |b| = ?

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.

 

[SclayP]

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
u = x^2du = 2x
So,

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

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:

-\frac{1}{2}\int^{1}_{-0.78}\frac{1}{1-v^2} \, dv

v = cos(u)
dv = -sen(u)

 

[Dick]

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.

 

[SclayP]

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 \frac{1}{2}\int\frac{1}{sen(u)} \, du which is the integral of csc(u) and looked it up on Wolfram and it gives me the funcion f(x) = ln(sen(\frac{u}{2}))-ln(cos(\frac{u}{2})) + C.

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

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

And so that is equal to \infty.

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

 

[Mark44]

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

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.

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

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

And so that is equal to \infty.

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

 

[SclayP]

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...

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

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

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

So that is equal to

-\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|\infty| = \infty

Thanks again...

 

[Mark44]

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...

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

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.

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

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

So that is equal to

-\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|\infty| = \infty

Thanks again...

 

[SclayP]

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
\int^{\frac{\pi}{2}}_{0}csc(x^2) \, du = -\frac{1}{2}ln(csc(x^2) + cot(x^2)) + C |^{\frac{\pi}{2}}_{0}

Then
\lim_{b\rightarrow 0} {-\frac{1}{2}ln|csc(x^2) + cot(x^2)| + C |^{\frac{\pi}{2}}_{b}}
\lim_{b \to 0} -\frac{1}{2}ln|1.60 - 1.25| + \frac{1}{2}ln|csc(b^2) + cot(b^2)|

 

[Mark44]

And it looks like both terms in the 2nd ln expression approach infinity.

 

[SclayP]

And it looks like both terms in the 2nd ln expression approach infinity.

So the integral does not converge after all...

 

[Dick]

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.