Definite Integral of ln*algebraic function

[maverick280857]

Hello...

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

$$I = \int_{0}^{1}\frac{x^2lnx}{\sqrt{1-x^2}}dx$$

Any suggestions?

Thanks and cheers
Vivek

 

[maverick280857]

Here's what I did:

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

$$J = lnx\int \frac{x^2}{\sqrt{1-x^2}}dx - \int \int \frac{x^2}{\sqrt{1-x^2}}dx \frac{dx}{x}$$

$$\int \frac{x^2}{\sqrt{1-x^2}}dx = -\int \frac{1-x^2}{\sqrt{1-x^2}}dx + \int \frac{1}{\sqrt{1-x^2}}dx = \frac{1}{2}(\sin^{-1}x-x\sqrt{1-x^2})$$

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

One of the integrals to be further computed is $$\int \frac{\sin^{-1}x}{x}dx$$. 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 $$\frac{-\pi}{8}(-1+ln4)$$ 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
$$\int \int dx \frac{dx}{x}$$
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.

$$I = \int \frac{x}{\sqrt{1-x^2}}(x\ln x) dx$$

Integration by parts, udv + vdu = uv

Let $$\frac{x}{\sqrt{1-x^2}}$$ be dv, and $$(x\ln x)$$ be u

Then v will be $$-(1 - x^2)^{\frac{1}{2}}$$

And du will be $$(1 + \ln x)$$

Now we need to integrate vdu wrt x.

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

The first term can be easily evaluated by making the substitution $x = \sin \theta$

Let's integrate the second term. Make the same substitution $x = sin \theta$.

The second term becomes $$-\cos^2(\theta) \ln(|\sin \theta|) d \theta$$

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

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

Integrating $$\sin (2\theta)d\theta$$ is trivial.

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

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

 

[dextercioby]

Hello...

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

$$I = \int_{0}^{1}\frac{x^2lnx}{\sqrt{1-x^2}}dx$$

Any suggestions?

Thanks and cheers
Vivek

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

The integral becomes
$$J=\int_{0}^{\frac{+\pi}{2}} \sin^{2}u \ \ln\sin u \ du$$

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.

 

[maverick280857]

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

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:

$$\int udv = uv - \int vdu$$

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

$$\int udv = u(\int f(x)dx) - \int (\int f(x)dx)du$$

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),

$u = lnx$
$dv = \frac{x^2}{\sqrt{1-x^2}}$ or equivalently $v = \int \frac{x^2}{\sqrt{1-x^2}}dx$

And by the way hallsofivy,

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

makes no sense at all- you can't do a double integral over the SAME variable!

Is that wrong? Just because you can find $\int a dt$ 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 $u = \frac{x^2}{\sqrt{1-x^2}}$, and $dv = lnx dx$ 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 $$\theta \cot \theta$$ which I cannot integrate easily. It still comes back to $$\int \ln|\sin \theta| d\theta$$ which I don't think has an elementary integral.

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

 

[maverick280857]

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 $$\theta \cot \theta$$ which I cannot integrate easily. It still comes back to $$\int \ln|\sin \theta| d\theta$$ which I don't think has an elementary integral.

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

Interestingly, I tried to do this problem in 3 ostensibly different ways on the test but each time I came up with $\int \theta \cot\theta d\theta$ and eventually $\int ln \sin\theta d\theta$. And that's 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
$I = \int^{1}_{0} \frac{x^{2} \ln x}{\sqrt{1-x^{2}}} \ dx \ .$

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.
$I = - \int^{1}_{0} x \ln x \cdot \frac{-x}{\sqrt{1-x^{2}}} \ dx = - \int^{1}_{0} x \ln x \cdot \frac{d}{dx}\sqrt{1-x^{2}} \ dx$
$= - \left[ x \ln x \sqrt{1-x^{2}} \right]^{1}_{0} + \int^{1}_{0} \left( 1 + \ln x \right) \sqrt{1-x^{2}} \ dx \ .$

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

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

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

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

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

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

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

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

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

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

 

[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 $\int^{\pi}_{0} \ln(\sin t) \ dt$ using complex integrals. Functions like $\ln z$ 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.

 

[asvani]

By the way, there's no closed form for the indefinite integral $\int \ln(\sin t) \ dt$. You'll end up with a series of horrible beta functions.

 

[dextercioby]

Okay.I don't know how u end up with beta functions,my Maple didn't...

Daniel.

 

[Zurtex]

Can someone please explain to me how the original integral is valid at x=0?

 

[dextercioby]

The original function has finite limits both to 0 and to 1...

Daniel.

 

[Curious3141]

Asvani, that was a *brilliant* solution ! I wish I'd thought of that !