# Thread: integral a la the Kriz way.

1. ## integral a la the Kriz way.

Here is an integral from the site that isomorphism posted. I thought I would try my hand at this one using Kriz's double integration technique.

$\displaystyle \int_{0}^{1}\frac{ln(x)}{1+x}dx=\frac{-{\pi}^{2}}{12}$

We know that $\displaystyle ln(x)=\int_{1}^{x}\frac{1}{u}du$

Then we get:

$\displaystyle \int_{0}^{1}\int_{1}^{x}\frac{1}{u(x+1)}dudx$

Now, here is where I am hung up on switching the limits of integration, coming up with the new limits, and thus coming up with a nice workable integral.

It looks similar to $\displaystyle \int_{0}^{1}\frac{ln(x+1)}{x^{2}+1}dx=\frac{\pi}{8 }ln(2)$

2. Well in that case, it won't work.

To reverse integration order, we rewrite the double integral as $\displaystyle -\int_{0}^{1}\!{\int_{x}^{1}{\frac{du\,dx}{u(x+1)}} }.$

Now play with inequalities: $\displaystyle 0\le x\le1,\,x\le u\le1.$ Hence $\displaystyle 0\le x\le u,\,0\le u\le1.$ After reversing integration order the double integral equals $\displaystyle -\int_{0}^{1}\!{\int_{0}^{u}{\frac{dx\,du}{u(x+1)}} },$ which is doable. (Only the inner one.)

But we have a problem, that will yield another undoable integral which result is exactly $\displaystyle -\frac{\pi^2}{12},$ so the only way to tackle this, it's a power series application.

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If you want to try a double integral trick, see $\displaystyle \int_{0}^{\pi /2}{\frac{\ln (\sec x)}{\tan x}\,dx}.$ You can watch my solution here.

3. Originally Posted by galactus
Here is an integral from the site that isomorphism posted. I thought I would try my hand at this one using Kriz's double integration technique.

$\displaystyle \int_{0}^{1}\frac{ln(x)}{1+x}dx=\frac{-{\pi}^{2}}{12}$

We know that $\displaystyle ln(x)=\int_{1}^{x}\frac{1}{u}du$

Then we get:

$\displaystyle \int_{0}^{1}\int_{1}^{x}\frac{1}{u(x+1)}dudx$

Now, here is where I am hung up on switching the limits of integration, coming up with the new limits, and thus coming up with a nice workable integral.

It looks similar to $\displaystyle \int_{0}^{1}\frac{ln(x+1)}{x^{2}+1}dx=\frac{\pi}{8 }ln(2)$
$\displaystyle \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n$

So we have

$\displaystyle \sum_{n=0}^{\infty}(-1)^n\int_0^{1}\ln(x)x^ndx$

Now

$\displaystyle \int_0^{1}\ln(x)x^n=\frac{-1}{(n+1)^2}$

So we have

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

Now we know that

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

So we have( Lets hope this works)

$\displaystyle -(1-2^{1-2})\zeta(2)=-\left(\frac{1}{2}\right)\frac{\pi^2}{6}=\frac{-\pi^2}{12}$

Hooray!

4. I dont know how to do it using double integrals. I always have a tough time switching the order of integration.

I have a series way of doing it:

For my ease, I started with $\displaystyle x = e^t$

So $\displaystyle I = \int_0^1 \frac{\ln x}{1+x} dx = \int_{-\infty}^0 \frac{t e^t}{1+e^t} dt =$$\displaystyle -\int_{-\infty}^0 \left(\sum_{n=1}^{n = \infty} t (-e^t)^n\right) dt Now switching integration and summation, \displaystyle I = -\sum_{n=1}^{n = \infty}(-1)^n \left(\int_{-\infty}^0 t e^{nt}\right) dt \displaystyle \int_{-\infty}^0 t e^{nt} \, dt = \frac{te^{nt}}{n}\bigg{|}_{-\infty}^0 - \frac{e^{nt}}{n^2}\bigg{|}_{-\infty}^0 = 0 - \frac1{n^2} Thus: \displaystyle I = -\sum_{n=1}^{n = \infty}(-1)^n \left(\int_{-\infty}^0 t e^{nt}\right) dt$$\displaystyle = \sum_{n=1}^{n = \infty} \frac{(-1)^n}{n^2}$

$\displaystyle = -\left(\sum_{n=1}^{n = \infty}\frac1{n^2} - \frac12 \sum_{n=1}^{n = \infty} \frac1{n^2}\right)$

$\displaystyle =- \left(\frac{\pi^2}6 - \frac{\pi^2}{12}\right) = -\frac{\pi^2}{12}$

5. Thanks for all the replies, kids. I was thinking it may not work, but thought it was me.

Actually, I worked it mathstud's way and got it OK because I recognized the $\displaystyle \frac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^{k}x^{k}$. I was just wondering about that double integral thing.