Definite Integral of ln*algebraic function

maverick280857

Hello...

I got a simple looking integral to solve but unfortunately couldn't do it

[tex]I = \int_{0}^{1}\frac{x^2lnx}{\sqrt{1-x^2}}dx[/tex]

Any suggestions?

Thanks and cheers
Vivek

maverick280857

Here's what I did:

Let [itex]J = \int \frac{x^2lnx}{\sqrt{1-x^2}}dx[/itex]. Integrating J by parts,

[tex]J = lnx\int \frac{x^2}{\sqrt{1-x^2}}dx - \int \int \frac{x^2}{\sqrt{1-x^2}}dx \frac{dx}{x}[/tex]

[tex]\int \frac{x^2}{\sqrt{1-x^2}}dx = -\int \frac{1-x^2}{\sqrt{1-x^2}}dx + \int \frac{1}{\sqrt{1-x^2}}dx = \frac{1}{2}(\sin^{-1}x-x\sqrt{1-x^2})[/tex]

So [tex]J = lnx(\frac{1}{2}(sin^{-1}x-x\sqrt{1-x^2}+) - \frac{1}{2}\int \frac{\sin^{-1}x-x\sqrt{1-x^2}}{x}dx[/tex]

One of the integrals to be further computed is [tex]\int \frac{\sin^{-1}x}{x}dx[/tex]. This leads us nowhere

I can't think of any property of definite integrals (thanks to the ln function) which I can apply here. I know that this is an improper integral as such so I'll have to take limits after I've evaluated the corresponding indefinite integral somehow.

Please help.....

Mathematica gives the answer as [tex]\frac{-\pi}{8}(-1+ln4)[/tex] but I had to do this by hand...

Thanks and cheers
Vivek

K.J.Healey

in your second line of equations with the double integral, are you sure you can bring that 1/x inside? I didnt think you could do that.

HallsofIvy

That's NOT integration by parts! and
[tex]\int \int dx \frac{dx}{x}[/tex]
makes no sense at all- you can't do a double integral over the SAME variable!

Curious3141

I've thought of a way you can evaluate the indef. integral. I haven't written out the working, but I'll give you the gist. Be warned, it's tedious. But fairly elementary.

[tex]I = \int \frac{x}{\sqrt{1-x^2}}(x\ln x) dx[/tex]

Integration by parts, udv + vdu = uv

Let [tex]\frac{x}{\sqrt{1-x^2}}[/tex] be dv, and [tex](x\ln x)[/tex] be u

Then v will be [tex]-(1 - x^2)^{\frac{1}{2}}[/tex]

And du will be [tex](1 + \ln x)[/tex]

Now we need to integrate vdu wrt x.

vdu can be expressed as the sum of [tex]-(1 - x^2)^{\frac{1}{2}}dx[/tex] and [tex]-(1 - x^2)^{\frac{1}{2}}(\ln x)dx[/tex]

The first term can be easily evaluated by making the substitution [itex]x = \sin \theta[/itex]

Let's integrate the second term. Make the same substitution [itex]x = sin \theta[/itex].

The second term becomes [tex]-\cos^2(\theta) \ln(|\sin \theta|) d \theta[/tex]

Integrate that by parts. Use [tex]\cos^2(\theta) = dv[/tex] and [tex] \ln(|\sin \theta|) = u[/tex]

Then vdu becomes [tex]-\frac{1}{2} \left( \cot \theta \cos(2\theta) + \cot \theta \right)d\theta[/tex] which can be simplified to [tex](\frac{1}{2}\sin 2\theta - \cot \theta)d \theta[/tex]

Integrating [tex]\sin (2\theta)d\theta[/tex] is trivial.

Integrating [tex]\cot \theta[/tex] is easy because, being the ratio of cosine to sine, it is of the form [tex]f'(x)g(f(x))[/tex] giving the integral as [tex]\ln|\sin\theta|[/tex]

Put everything together and you'll have your answer as an indefinite integral. Evaluate for the bounds for the final numerical answer.

dextercioby

I don't know what mathematica did to get that answer. Try the substitution
[tex] x=\sin u [/tex]
under which the limits of integration become
[tex] u_{1}=0;u_{2}=\frac{\pi}{2} [/tex]

The integral becomes
[tex] J=\int_{0}^{\frac{+\pi}{2}} \sin^{2}u \ \ln\sin u \ du [/tex]

Mathematica on the wolfram's site gives the antiderivative:
(see attachement) which is pretty nasty.Not to mention the fact that applying the FTC will produce a terrible headache.

Daniel.

Attached Thumbnails

 

maverick280857

First of all, you folks are interpreting the integral incorrectly. I never said you could take the x inside!! The first integral must be evaluated first and then the second one:

[tex]\int udv = uv - \int vdu[/tex]

In my case v is a function defined by an integral (say [itex]v=\int f(x)dx[/itex]). So the above integral can be rewritten in terms of f(x) as:

[tex]\int udv = u(\int f(x)dx) - \int (\int f(x)dx)du[/tex]

It is obviously understood that the two integrals cannot be commuted. So I thought you would understand and hence did not place the parantheses. Sorry for the confusion though.

Also when I entered the original integral in Mathematica with the limits, I got the answer mentioned in my second post. I wonder how you got a different answer dextercioby. Curious3141, I'm going to try out the approach you've suggested in a while. Thanks for your help mates

I did make the obvious substitution but couldn't take it further by hand. (This was a question on a math test ).

Cheers
Vivek

maverick280857

To add to my last post (to clear your doubts Healey01 and Hallsofivy),

[itex]u = lnx[/itex]
[itex]dv = \frac{x^2}{\sqrt{1-x^2}}[/itex] or equivalently [itex]v = \int \frac{x^2}{\sqrt{1-x^2}}dx[/itex]

And by the way hallsofivy,

velocity = [itex]v = \frac{dx}{dt}[/tex]
acceleration = [itex]a = \frac{dv}{dt} = \frac{d^2x}{dt^2}[/itex]
position = [itex]x = \int v dt = \int (\int a dt)dt[/itex]

Is that wrong? Just because you can find [itex]\int a dt[/itex] first and call it v and then put it inside the first integral doesn't mean you can't write it this way?!????

vincentchan

let your [itex] u = \frac{x^2}{\sqrt{1-x^2}}[/itex], and [itex] dv = lnx dx [/itex] and do it by part....

Curious3141

Vivek, I think I may have made a bad error in my computation in the later steps. I tried it again, and I get an integrand of [tex]\theta \cot \theta[/tex] which I cannot integrate easily. It still comes back to [tex]\int \ln|\sin \theta| d\theta[/tex] which I don't think has an elementary integral.

Sorry I misled you. :( I think this isn't an elementary integral at all.

maverick280857

Interestingly, I tried to do this problem in 3 ostensibly different ways on the test but each time I came up with [itex]\int \theta \cot\theta d\theta[/itex] and eventually [itex]\int ln \sin\theta d\theta[/itex]. And thats when I gave up so when I saw your first post, I thought I was indeed missing something....

cheers
vivek

vincentchan

why don't you use my method posted above.....

dextercioby

I don't know,because it lead...nowhere...?

Da

asvani

The definite integral to be done is given by
[latex]
I = \int^{1}_{0} \frac{x^{2} \ln x}{\sqrt{1-x^{2}}} \ dx \ .
[/latex]

The primary difficulty of this integral is that it contains a logarithmic function and a square root in the integrand. However, it can be simplified by an integration of parts.
[latex]
I
= - \int^{1}_{0} x \ln x \cdot \frac{-x}{\sqrt{1-x^{2}}} \ dx
= - \int^{1}_{0} x \ln x \cdot \frac{d}{dx}\sqrt{1-x^{2}} \ dx
[/latex]
[latex]
= - \left[ x \ln x \sqrt{1-x^{2}} \right]^{1}_{0}
+ \int^{1}_{0} \left( 1 + \ln x \right) \sqrt{1-x^{2}} \ dx \ .
[/latex]

It can be easily determined that the first term on the right of the above equation is zero. Hence, we have
[latex]
I
= \int^{1}_{0} \sqrt{1-x^{2}} \ dx + \int^{1}_{0} \ln x \sqrt{1-x^{2}} \ dx \ .
[/latex]

The first term on the RHS can be easily solved by using the substitution [latex]x=\sin\theta[/latex], which yields
[latex]
\int \sqrt{1-x^{2}} \ dx
= \frac{1}{2} \left( \sin^{-1}x + x\sqrt{1-x^{2}} \right) \ .
[/latex]

Thus, we have
[latex]
I
=
\frac{1}{2} \left[ \sin^{-1}x + x\sqrt{1-x^{2}} \right]^{1}_{0} +
\int^{1}_{0} \ln x \sqrt{1-x^{2}} \ dx
=
\frac{\pi}{4} + \int^{1}_{0} \ln x \sqrt{1-x^{2}} \ dx \ .
[/latex]

The second term can be simplified by applying another integration by part.
[latex]
I
=
\frac{\pi}{4} + \int^{1}_{0} \ln x \sqrt{1-x^{2}} \ dx
\nonumber \\
=
\frac{\pi}{4} +
\frac{1}{2} \int^{1}_{0} \ln x \cdot
\frac{d}{dx}\left( \sin^{-1}x + x\sqrt{1-x^{2}} \right) \ dx
\nonumber \\
[/latex]
[latex]
=
\frac{\pi}{4} +
\frac{1}{2} \left[\left(\sin^{-1}x + x\sqrt{1-x^{2}}\right)\right]^{1}_{0} -
\frac{1}{2} \int^{1}_{0} \frac{1}{x} \left(\sin^{-1}x + x\sqrt{1-x^{2}}\right) \ dx
\nonumber \\
=
\frac{\pi}{8} - \frac{1}{2} \int^{1}_{0} \frac{\sin^{-1}x}{x} \ dx \ .
[/latex]

The integral [latex]I[/latex] is now more tractable now that we have got rid of the [latex]\log[/latex] function. The solution can be completed once we evaluate the integral term in the last eqation. Using the substitution [latex]x=\sin t[/latex] and applying an integration by part,
[latex]
\int^{1}_{0} \frac{\sin^{-1}x}{x} \ dx
=
\int^{\pi/2}_{0} t \cot t \ dt
=
\int^{\pi/2}_{0} t \frac{d}{dt}\ln(\sin t) \ dt
[/latex]
[latex]
=
\left[t \ln(\sin t)\right]^{\pi/2}_{0} -
\int^{\pi/2}_{0} \ln(\sin t) \ dt
=
- \int^{\pi/2}_{0} \ln(\sin t) \ dt \ .
[/latex]

Thus, it simplifies to
[latex]
I
= \frac{\pi}{8} + \frac{1}{2} \int^{\pi/2}_{0} \ln(\sin t) \ dt \ .
[/latex]

Fortunately, the definite integral term in the above equation can be readily evaluated.
[latex]
\int^{\pi/2}_{0} \ln(\sin t) \ dt
=
\frac{1}{2} \int^{\pi}_{0} \ln(\sin t) \ dt
=
\frac{1}{2} \int^{\pi}_{0} \ln\left(2\sin(t/2)\cos(t/2)\right) \ dt
[/latex]
[latex]
=
\frac{1}{2} \int^{\pi}_{0} \ln 2 \ dt +
\frac{1}{2} \int^{\pi}_{0} \ln(\sin(t/2)) \ dt +
\frac{1}{2} \int^{\pi}_{0} \ln(\cos(t/2)) \ dt
=
\frac{\pi}{2} \ln 2 + 2 \int^{\pi/2}_{0} \ln(\sin t) \ dt
[/latex]

since
[latex]
\frac{1}{2} \int^{\pi}_{0} \ln(\sin(t/2)) \ dt = \int^{\pi/2}_{0} \ln(\sin t) \ dt
[/latex]
and
[latex]
\frac{1}{2} \int^{\pi}_{0} \ln(\cos(t/2)) \ dt
= \int^{\pi/2}_{0} \ln(\cos t) \ dt
= \int^{\pi/2}_{0} \ln(\sin t) \ dt \ .
[/latex]

Therefore, it is obvious that
[latex]
\int^{\pi/2}_{0} \ln(\sin t) \ dt
= -\frac{\pi}{2} \ln 2
[/latex]

and that the final answer is
[latex]
I = \frac{\pi}{8}\left(1-\ln 4\right) \ .
[/latex]

asvani

My answer is consistent with Mathematica's so it's alright!

dextercioby

Yes,i didn't claim it could't be done (congratulations!!),just that it was very difficult to find the antiderivative and the apply the FTC...You made use however of the fact that it was a definite integral...

Daniel.

asvani

My first stab three weeks ago at evaluating that integral ended up horribly as I attempted to evaluate [latex]\int^{\pi}_{0} \ln(\sin t) \ dt[/latex] using complex integrals. Functions like [latex]\ln z[/latex] are neither very analytic nor friendly with branch cuts appearing in unwanted places. I gave up after trying for an hour. Then, I had to report to camp for reservist training for the next three weeks...

In the course of doing the complex integrals, I fiddled around with the limits of the integral and discovered some nice symmetries in the definite integral. After that, it was all downhill.