or where do I learn how to do the introducing two integrals to eliminate the complexity of an integral? I mean how do you do it? Can someone spare a minute to fill me in?

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- Apr 1st 2008, 06:21 PM #1

- Apr 1st 2008, 06:55 PM #2
I have to say that I really love that method, I personally use it in a variety of cases, but not always works.

For example, compute $\displaystyle \int_1^2\ln x\,dx.$ The straightforward sense tells us that we must apply integration by parts, but look another way. For $\displaystyle x>0,\,\ln x=\int_1^x\frac1y\,dy,$ and there it is, we have our parameter to construct our double integral with the only reason to reverse integration order and make a little bit simple the problem.

Another example, find $\displaystyle \int_0^1\ln(1+\sqrt x)\,dx.$ Again, follow the same idea, construct a double integral and reverse integration order, you'll get a simple integral to tackle.

How about the famous Dirichlet Integral $\displaystyle \int_0^\infty\frac{\sin x}{x}\,dx$? This can also be tackled by introducing $\displaystyle \frac1x=\int_0^\infty e^{-xy}\,dy,\,x>0,$ the result follows... and so on.

A lot of single integrals can be solved with double integration tricks, but it depends.

- Apr 2nd 2008, 12:23 PM #3

- Apr 2nd 2008, 02:44 PM #4

- Apr 2nd 2008, 04:58 PM #5

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As you can see Krizalid likes to use double integrals, which really help a lot. I like to use infinite series. But those can be hard to sum at times. Complex analysis also helps, but again not always. You should learn different techniques for integration just in case on approach is really bad.

Here is another example. I was talking with my classmates today about integration tricks and I mentioned double integration.

Consider, $\displaystyle \int_0^{\infty} \frac{\cos (bx) - \cos (ax)}{x^2} dx = \int_0^{\infty} \int_b^a \frac{\sin (yx)}{x} dy ~ dx$.

Now change order of integration and you will get, $\displaystyle \frac{\pi}{2}(b-a)$ as the answer.

- Apr 2nd 2008, 05:07 PM #6

- Apr 2nd 2008, 05:10 PM #7

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- Apr 2nd 2008, 05:15 PM #8
## I thought I did

Dont you evaluate just evaluate them as normal one at a time and regard the x or y that isnt the integration variable(dx,dy) as a constant then integrate the other?

And also how do you get them into these new forms I understood the $\displaystyle \frac1x=\int_0^\infty e^{-xy}\,dy,\,x>0,$ but I dont understand how you converted your example...nor do I understand the Dirichlet application

- Apr 2nd 2008, 07:48 PM #9

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Exactly. But what the trick is to change the order of integration instead of integrating (dx,dy) you integrate (dy,dx).

And also how do you get them into these new forms I understood the $\displaystyle \frac1x=\int_0^\infty e^{-xy}\,dy,\,x>0,$ but I dont understand how you converted your example...nor do I understand the Dirichlet application

- Apr 3rd 2008, 12:38 PM #10
## Well

$\displaystyle \int_1^2\ln x\,dx\$=$\displaystyle \int_1^2\int_1^{x}\ln y\,dy\$ so you get ]=$\displaystyle \int_1^{x}\int_1^2\frac{1}{y}\,dy\$=$\displaystyle \int_1^{x}\ln(2)-ln(1)\$=$\displaystyle \int_1^{x}\ln(2))\$=$\displaystyle \ln(2)x-ln(2)\$

but I must have done something wrong because that doesnt make sense?

- Apr 3rd 2008, 12:39 PM #11

- Apr 3rd 2008, 12:49 PM #12

- Apr 3rd 2008, 01:05 PM #13
## Ok is this semi-right?

$\displaystyle \int_1^2\ln x\,dx\$=$\displaystyle \int_1^2\int_1^{x}\frac{1}{y}\,dy,dx\$=]=$\displaystyle \int_1^{x}\int_1^2\frac{1}{y}\,dx,dy\$=$\displaystyle \int_1^{x}\frac{2}{y}-\frac{1}{y}\,dy\$=ln(x) what am I doing wrong will you point it out to me?

- Apr 3rd 2008, 06:36 PM #14

- Apr 3rd 2008, 07:14 PM #15