cool integral

cody · **Joined:** Sun, 26 Sep 2004 23:12:55 UTC **Posts:** 408

Give this one a go:

$\int_{0}^{1}\int_{0}^{1}\frac{1}{ln(xy)(1+(xy)^{2})}dxdy$

commutative · **Joined:** Sat, 18 Mar 2006 08:42:24 UTC **Posts:** 834

cody wrote:

Give this one a go:

$$ I=\int_{0}^{1}\int_{0}^{1}\frac{1}{\ln(xy)(1+(xy)^{2})}dxdy$

Spoiler:

cody · **Joined:** Sun, 26 Sep 2004 23:12:55 UTC **Posts:** 408

Thanks for the response, commutative.

Actually, that is basically how I approached it, except I used u instead of v as a sub. I reckon I won't bother posting my result then. At least now I know I done it correctly.

The original problem was multiplied by -4, therefore, resulting in Pi as the solution.

What I find amazing about these types of integrals is that one can run them through Maple or Mathematica and not get a result( if you do get a result it is a long complicated thing), yet human ingenuity easily pumps out the answer.

I ran this one through Maple and I couldn't get it to give me a solution, rather definite or indefinite.

commutative · **Joined:** Sat, 18 Mar 2006 08:42:24 UTC **Posts:** 834

here's a "cool" integral: $$ \int_0^{\frac{\pi}{2}} x\ln(\sin x) \ dx = ?$

Hint:

Spoiler:

cody · **Joined:** Sun, 26 Sep 2004 23:12:55 UTC **Posts:** 408

Hey commutative. That is a cool integral.

I haven't had much time to spend with it, but here are a few quick thoughts I had.

Using parts, I got it to:

$\frac{-1}{2}\int_{0}^{\frac{\pi}{2}}x^{2}cot(x)dx$

Now, this has a series of $x^{2}cot(x)=\frac{-1}{2}\sum_{n=0}^{\infty}(-4)^{n}\frac{B_{2n}}{(2n)!}x^{2n+1}$

Where $B_{n}$ are the Bernoulli numbers. But what to do with that thing now?.

It would be easier to use $\int_{0}^{\frac{\pi}{2}}\int_{0}^{cot(x)}x^{2}dydx$

I also thought about $\int_{1}^{sin(x)}\frac{x}{y}dy=xln(sin(x))$

That is as far as I have gotten for now. What is your approach?. I bet it is clever, whatever it is. Parhaps you can show me a few tricks and cool identities to use.

Also, if we make the sub u=sin(x)

, we get:

$\int_{0}^{1}\frac{sin^{-1}(u)ln(u)}{\sqrt{1-u^{2}}}du$

This looks nasty, but it may be easier to work with than the others I mentioned.

We can use the identity $\frac{sin^{-1}(u)}{\sqrt{1-u^{2}}}=\sum_{n=0}^{\infty}\frac{2^{2n}(n!)^{2}}{(2n+1)!}x^{2n+1}$

Just brainstorming for now.

cody · **Joined:** Sun, 26 Sep 2004 23:12:55 UTC **Posts:** 408

Well, here is what I managed to do, but I got hung up.

After making the sub $u=sin(x), \;\ x=sin^{-1}(u), \;\ dx=\frac{1}{\sqrt{1-u^{2}}}du$

$\int_{0}^{1}\frac{sin^{-1}(u)ln(u)}{\sqrt{1-u^{2}}}du$

Then, using the aforementioned identity:

$\int_{0}^{1}\sum_{n=0}^{\infty}\frac{2^{2n}ln(u)(n!)^{2}}{(2n+1)!}u^{2n+1}$

But, $\int_{0}^{1}u^{2n+1}ln(u)du=\frac{-1}{4(n+1)^{2}}$

$\frac{-1}{4}\sum_{n=0}^{\infty}\frac{2^{2n}(n!)^{2}}{(2n+1)!(n+1)^{2}}$

Now, using gamma:

$\frac{-1}{4}\sum_{n=0}^{\infty}\frac{2^{2n}n{\Gamma}(n+1){\Gamma}(n)}{(n+1)^{2}{\Gamma}(2n+1)}$

But this equals B(n+1, n+1)

, so I used the beta integral and got:

$\frac{-1}{4}\int_{0}^{1}\sum_{n=0}^{\infty}\frac{2^{2n}}{(n+1)^{2}}x^{n}(1-x)^{n}$

Now, I got stuck. Perhaps you can help me finish or show me the error of my ways. I am sure there is an easier way, but this looked promising until I got hung up on the last part. Is there some little thing I am not seeing?. Is there a ${\zeta}(3)$ in there somewhere?.

cody · **Joined:** Sun, 26 Sep 2004 23:12:55 UTC **Posts:** 408

I tried editing my post, but it didn't take.

Here are some other thoughts I had.

We could use the series for ln(1-x).

Since $sin(x)=\sqrt{1-cos^{2}(x)}$ , we can write:

$xln(\sqrt{1-cos^{2}(x)})=\frac{x}{2}ln(1-cos^{2}(x))$

This gives the series:

$\frac{-1}{2}\sum_{n=1}^{\infty}\frac{xcos^{2n}(x)}{n}$

Another is to use $\frac{-1}{2}\int_{0}^{1}\frac{(sin^{-1}(x))^{2}}{x}dx$

Then use the series and go from there.

Mathstud28 · **Posted:** Wed, 30 Jul 2008 16:57:36 UTC

Sorry for not getting back to you on MHF Galctus...I have been up for 39 hours and I jsut forgot! Sorry...but let me pose this question for Commutative...do you mean to solve this by use of the power series of arcsine squared with reciprocated central binomial coefficients etc? Or is there an easier way?

cody · **Joined:** Sun, 26 Sep 2004 23:12:55 UTC **Posts:** 408

Whatever way you want to buddy. Using the series and gamma/beta thing would be nice. But whatever. There are many ways to go about it I have found. I get close but not quite. I listed several here and on MHF

WE can even use the series for ln(1-x). Or $\frac{-1}{2}\int_{0}^{1}x^{2}cot(x)dx$

cody · **Joined:** Sun, 26 Sep 2004 23:12:55 UTC **Posts:** 408

Well, it doesn't appear anyone is interested anyway, but I finally found a solution to this infernal thing.

Using $sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$

Factor:

$\frac{e^{ix}(1-e^{-2ix})}{2i}$

Now, if we use the series for ln(1-a), we get $xln(1-e^{-2ix})$

Plus some other stuff to integrate because of the $\frac{e^{ix}}{2i}$ .

This all leads us to the solution:

$\boxed{\frac{7}{16}{\zeta}(3)-\frac{{\pi}^{2}}{8}ln(2)}$

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