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send me the full solution to solve this!!!!!!!!

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- Sep 20th 2007, 07:20 PMkamaksh_iceimproper arctangent integral
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send me the full solution to solve this!!!!!!!! - Sep 20th 2007, 07:23 PMThePerfectHacker
- Sep 20th 2007, 07:25 PMJhevon
- Sep 20th 2007, 08:42 PMThePerfectHacker
Amazingly this integral appears in my Advanced Calculus book so the credit is

**not mine**(though I wish it was).

First you must say $\displaystyle 0<b<a$. Confirm it is uniformly continous and do the following double integral trick.

$\displaystyle \int_0^{\infty} \frac{\tan^{-1} ax - \tan^{-1} bx}{x} dx = \int_0^{\infty} \int_b^a \frac{1}{t^2 x^2 + 1} \ dt \ dx$.

Change order,

$\displaystyle \int_b^a \int_0^{\infty} \frac{1}{t^2x^2+1} \ dx \ dt = \int_b^a \frac{\tan^{-1} tx}{t} \big|_0^{\infty} \ dx = \int_a^b \frac{\pi}{2t} dt = \frac{\pi}{2} \ln \frac{a}{b}$ - Sep 20th 2007, 09:19 PMKrizalid
More generally

If $\displaystyle f'(x)$ is continuous and the integral converges,

$\displaystyle \int_0^\infty\frac{f(ax)-f(bx)}x\,dx=(f(0)-f(\infty))\ln\frac ba$

This beautiful result is called Frullani's Integral.

Nice, isn't it? - Sep 21st 2007, 07:01 AMThePerfectHacker
Yes it is nice (Clapping). Where did you get that from? It seems MathWorld has it while Wikipedia does not.

You should also state the conditions on the function* and limits. Perhaps $\displaystyle a>b>0$.

*)Maybe it should also be uniformly continous and having a limit at infinity? - Dec 7th 2007, 06:28 PMKrizalid
I found (not mine) a proof of Frullani's Integral, and I'd like to share it :)

Let

$\displaystyle \int_\varphi ^\tau {\frac{{f(ax) - f(bx)}}

{x}\,dx}.$

So

$\displaystyle \int_\varphi ^\tau {\frac{{f(ax) - f(bx)}}

{x}\,dx} = \int_\varphi ^\tau {\frac{{f(ax)}}

{x}\,dx} - \int_\varphi ^\tau {\frac{{f(bx)}}

{x}\,dx} = \int_{a\varphi }^{a\tau } {\frac{{f(x)}}

{x}\,dx} - \int_{b\varphi }^{b\tau } {\frac{{f(x)}}

{x}\,dx} .$

And

$\displaystyle \int_\varphi ^\tau {\frac{{f(ax) - f(bx)}}

{x}\,dx} = \int_{a\varphi }^{b\varphi } {\frac{{f(x)}}

{x}\,dx} - \int_{a\tau }^{b\tau } {\frac{{f(x)}}

{x}\,dx} = \int_a^b {\frac{{f(u\varphi )}}

{u}\,du} - \int_a^b {\frac{{f(u\tau )}}

{u}\,du} .$

Now let $\displaystyle \varphi \to 0,\,\tau \to \infty ,$

$\displaystyle \int_a^b {\frac{{f(0)}}

{u}\,du} = f(0)\ln \frac{b}

{a}$ & $\displaystyle \int_a^b {\frac{{f(\infty )}}

{u}\,du} = f(\infty )\ln \frac{b}

{a}.$

The conclusion follows.