# Math Help - Another integral

1. ## Another integral

Ok how would you go about doing this one $\int_0^{\infty}\frac{arctan(ax)-arctan(bx)}{x}dx$?....I would imagine you go $\int_0^{\infty}\bigg[\sum_{n=0}^{\infty}\frac{(-1)^n\cdot{a^{2n+1}}\cdot{x^{2n}}}{2n+1}-\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{b^{2n+1}}\cdot{x^{2n}}}{2n+1}\bigg]{dx}$ $=\bigg[\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{a^{2n+1}}\cdot{x^{2n+1}}}{(2n+1)^2}-\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{b^{2n+1}}\cdot{x^{2n+1}}}{(2n+1)^2}\bi gg]\bigg|_0^{\infty}$

2. Originally Posted by Mathstud28
Ok how would you go about doing this one $\int_0^{\infty}\frac{arctan(ax)-arctan(bx)}{x}dx$?....I would imagine you go $\int_0^{\infty}\bigg[\sum_{n=0}^{\infty}\frac{(-1)^n\cdot{a^{2n+1}}\cdot{x^{2n}}}{2n+1}-\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{b^{2n+1}}\cdot{x^{2n}}}{2n+1}\bigg]{dx}$ $=\bigg[\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{a^{2n+1}}\cdot{x^{2n+1}}}{(2n+1)^2}-\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{b^{2n+1}}\cdot{x^{2n+1}}}{(2n+1)^2}\bi gg]\bigg|_0^{\infty}$
There are many possible approaches. Here's a simple one:

Let $f(a) = \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx$.

Then $f'(a) = \frac{d}{da} \left[\int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx\right] = \int_{0}^{\infty} \frac{d}{da} \left[ \frac{\tan^{-1}(ax)}{x} \right] \, dx$

$= \int_{0}^{\infty} \frac{1}{a^2 x^2 + 1} \, dx = \frac{1}{a} \, \tan^{-1}(ax) \bigg{|}_{0}^{\infty} = \frac{\pi}{2a}$.

Therefore $f(a) = \int \frac{\pi}{2a} \, da = \frac{\pi}{2} \ln (a) + C$.

Therefore $f(a) - f(b) = \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx - \int_{0}^{\infty} \frac{\tan^{-1}(bx)}{x} \, dx$

$= \left(\frac{\pi}{2} \ln (a) + C\right) - \left(\frac{\pi}{2} \ln (b) + C\right) = \frac{\pi}{2} \ln\left( \frac{a}{b}\right)$.

I'm sure the big K will outline the other approaches

3. Originally Posted by mr fantastic
There are many possible approaches. Here's a simple one:

Let $f(a) = \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx$.

Then $f'(a) = \frac{d}{da} \left[\int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx\right] = \int_{0}^{\infty} \frac{d}{da} \left[ \frac{\tan^{-1}(ax)}{x} \right] \, dx$

$= \int_{0}^{\infty} \frac{1}{a^2 x^2 + 1} \, dx = \frac{1}{a} \, \tan^{-1}(ax) \bigg{|}_{0}^{\infty} = \frac{\pi}{2a}$.

Therefore $f(a) = \int \frac{\pi}{2a} \, da = \frac{\pi}{2} \ln (a) + C$.

Therefore $f(a) - f(b) = \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx - \int_{0}^{\infty} \frac{\tan^{-1}(bx)}{x} \, dx$

$= \left(\frac{\pi}{2} \ln (a) + C\right) - \left(\frac{\pi}{2} \ln (b) + C\right) = \frac{\pi}{2} \ln\left( \frac{a}{b}\right)$.
That is a really neat method...but does that always work or does it work conveniently here...or could I always go $\int_a^{b}F(x)dx=\int[f(b)-f(a)]\bigg|_a^{b}dx$?

4. Originally Posted by Mathstud28
That is a really neat method...but does that always work or does it work conveniently here...or could I always go $\int_a^{b}F(x)dx=\int[f(b)-f(a)]\bigg|_a^{b}dx$?
Read this (scroll down to examples): Differentiation under the integral sign - Wikipedia, the free encyclopedia

The technique was used extensively by Feynman (you can read about that in this book: Amazon.com: Surely You're Joking, Mr. Feynman! (Adventures of a Curious Character): Richard P. Feynman,Ralph Leighton,Edward Hutchings,Albert R. Hibbs: Books)

It can always be used with definite integrals, but will not always be useful. Obviously it was useful in this case.

5. Originally Posted by mr fantastic
Read this (scroll down to examples): Differentiation under the integral sign - Wikipedia, the free encyclopedia

The technique was used extensively by Feynman (you can read about that in this book: Amazon.com: Surely You're Joking, Mr. Feynman! (Adventures of a Curious Character): Richard P. Feynman,Ralph Leighton,Edward Hutchings,Albert R. Hibbs: Books)

It can always be used with definite integrals, but will not always be useful. Obviously it was useful in this case.
Thanks a lot!

6. $\frac{{\arctan (\alpha x) - \arctan (\beta x)}}
{x} = \int_\beta ^\alpha {\frac{1}
{{1 + x^2 y^2 }}\,dy}.$
So,

$\int_0^\infty {\frac{{\arctan (\alpha x) - \arctan (\beta x)}}
{x}\,dx} = \int_0^\infty \! {\int_\beta ^\alpha {\frac{{dy\,dx}}
{{1 + x^2 y^2 }}} }.$
After reverse integration order and some straightforward calculations yields the answer.

7. Originally Posted by Krizalid
$\frac{{\arctan (\alpha x) - \arctan (\beta x)}}
{x} = \int_\beta ^\alpha {\frac{1}
{{1 + x^2 y^2 }}\,dy}.$
So,

$\int_0^\infty {\frac{{\arctan (\alpha x) - \arctan (\beta x)}}
{x}\,dx} = \int_0^\infty \! {\int_\beta ^\alpha {\frac{{dy\,dx}}
{{1 + x^2 y^2 }}} }.$
After reverse integration order and some straightforward calculations yields the answer.
Ok I see that

8. This is just because I can't remember, but what are the conditions for reversing the order of integration ?

9. Originally Posted by Moo
This is just because I can't remember, but what are the conditions for reversing the order of integration ?
easiest way is to graph the region of integration and change the functions of x to functions of y and vice versa. this will give you the new limits to change the integration order

10. I mean aren't there any condition on the function, such as $C^1$, derivability or whatever ?

11. Originally Posted by Moo
I mean aren't there any condition on the function, such as $C^1$, derivability or whatever ?
oh, well, in that case, i don't remember either . As far as reversing integration order goes, I think we worry about integrability, not differentiability. we would worry about differentiability when introducing the second integral in the first place.

12. Originally Posted by Moo
This is just because I can't remember, but what are the conditions for reversing the order of integration ?
Scroll down to theorem 9.4 (Fubini's Theorem - Stronger Form): Repeated Integrals and Fubini's Theorem

13. Ok, so only continuity !
Gosh... I've studied this theorem two weeks ago only :/

Thanks