# improper arctangent integral

• Sep 20th 2007, 07:20 PM
kamaksh_ice
improper arctangent integral
• Sep 20th 2007, 07:23 PM
ThePerfectHacker
Quote:

Originally Posted by kamaksh_ice
∫_0^∞((tan^(-1)ax-tan^(-1)bx) )/ax dx

Fix it!
• Sep 20th 2007, 07:25 PM
Jhevon
Quote:

or else what? (Dull)

Quote:

Originally Posted by kamaksh_ice
∫_0^∞((tan^(-1)ax-tan^(-1)bx) )/ax dx

what are these ? signs about?

did you mean: $\displaystyle \int_{0}^{\infty} \frac {\tan^{-1} ax - \tan^{-1} bx}{ax}~dx$ ?
• Sep 20th 2007, 08:42 PM
ThePerfectHacker
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 PM
Krizalid
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 AM
ThePerfectHacker
Quote:

Originally Posted by Krizalid
More generally

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

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

This beautiful result is called Frullani's Integral.

Nice, isn't it?

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 PM
Krizalid
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.