1. Originally Posted by NonCommAlg
3) $\displaystyle \int_0^{\infty}\cot^{-1}(ax) \cot^{-1}(bx) \ dx, \ \ a > 0, \ b > 0$
Have fune!

2. Originally Posted by pankaj
okay! start from the identity $\displaystyle \cot^{-1}(cx) = \int_c^{\infty} \frac{x}{x^2t^2 + 1} \ dt, \ x > 0, \ c > 0.$ therefore:

$\displaystyle \int_0^{\infty} \cot^{-1}(ax) \cot^{-1}(bx) \ dx = \int_0^{\infty} \int_a^{\infty} \int_b^{\infty} \frac{x^2}{(x^2y^2 + 1)(x^2z^2 + 1)} \ dy \ dz \ dx$

$\displaystyle =\int_a^{\infty} \int_b^{\infty} \int_0^{\infty} \frac{x^2}{(x^2y^2 + 1)(x^2z^2 + 1)} \ dx \ dy \ dz=\frac{\pi}{2} \int_a^{\infty} \int_b^{\infty} \frac{1}{yz(y+z)} \ dy \ dz$

$\displaystyle =\frac{\pi}{2} \int_a^{\infty} \frac{\ln(z+b) - \ln b}{z^2} \ dz = \frac{\pi}{2}\left[\frac{1}{a} \ln \left(\frac{a+b}{b} \right) + \frac{1}{b} \ln \left(\frac{a+b}{a} \right) \right].$

in the last integral i used integration by parts: $\displaystyle \ln(z+b) - \ln b = u, \ \ \frac{1}{z^2} dz = dv.$

3. Originally Posted by NonCommAlg
okay! start from the identity $\displaystyle \cot^{-1}(cx) = \int_c^{\infty} \frac{x}{x^2t^2 + 1} \ dt, \ x > 0, \ c > 0.$ therefore:

$\displaystyle \int_0^{\infty} \cot^{-1}(ax) \cot^{-1}(bx) \ dx = \int_0^{\infty} \int_a^{\infty} \int_b^{\infty} \frac{x^2}{(x^2y^2 + 1)(x^2z^2 + 1)} \ dy \ dz \ dx$

$\displaystyle =\int_a^{\infty} \int_b^{\infty} \int_0^{\infty} \frac{x^2}{(x^2y^2 + 1)(x^2z^2 + 1)} \ dx \ dy \ dz=frac{\pi}{2}\;\int_a^{\infty} \frac{\pi}{2} \int_a^{\infty} \int_b^{\infty} \frac{1}{yz(y+z)} \ dy \ dz$

$\displaystyle =\frac{\pi}{2} \int_a^{\infty} \frac{\ln(z+b) - \ln b}{z^2} \ dz = \frac{\pi}{2}\left[\frac{1}{a} \ln \left(\frac{a+b}{b} \right) + \frac{1}{b} \ln \left(\frac{a+b}{a} \right) \right].$

in the last integral i used integration by parts: $\displaystyle \ln(z+b) - \ln b = u, \ \ \frac{1}{z^2} dz = dv.$
Here is another method to solve the integral.

$\displaystyle F(a,b)\;=\;\int_0^\infty \cot^{-1}(ax) \cot^{-1}(bx)\; dx$

$\displaystyle \frac{\partial^2 F}{\partial a \; \partial b}\;=\;\int_0^\infty \frac{x^2}{(a^2x^2+1)(b^2x^2+1)} \; dx\;=\frac{\pi}{2} \frac{1}{ab(a+b)}$

$\displaystyle \frac{\partial F}{\partial a}\;=\; \frac{\pi}{2a} \int_\infty^b \frac{1}{b(a+b)}\;db \;=\; \frac{\pi}{2a^2}\ln\frac{a}{a+b}$

$\displaystyle F(a,b)\;=\; \; \frac{\pi}{2} \int_\infty^a \frac{1}{a^2}\ln\frac{a}{a+b}\, da\;=\;\frac{\pi}{2}\left[\frac{1}{a} \ln \left(\frac{a+b}{b} \right) + \frac{1}{b} \ln \left(\frac{a+b}{a} \right) \right]$

We integrate from $\displaystyle \infty$ because when the integral is 0 when either $\displaystyle a\text{ or } b \text{ is } \infty$

4. Originally Posted by luobo
Here is a simpler and easier method.

$\displaystyle F(a,b)\;=\;\int_0^\infty \cot^{-1}(ax) \cot^{-1}(bx)\; dx$

$\displaystyle \frac{\partial^2 F}{\partial a \; \partial b}\;=\;\int_0^\infty \frac{ab}{(a^2x^2+1)(b^2x^2+1)} \; dx=\frac{\pi}{2} \frac{ab}{a+b}$

Then integrate to get $\displaystyle F(a,b)$ .
i think your differentiation is not correct! anyway, these two methods are actually the same. the only difference is that i personally don't like differentiating with respect to a parameter in the

integrand. i'm not sure why though! haha

5. Originally Posted by NonCommAlg
i think your differentiation is not correct! anyway, these two methods are actually the same. the only difference is that i personally don't like differentiating with respect to a parameter in the

integrand. i'm not sure why though! haha
IT IS CORRECT. It turns out to be so simple that it may have surprised you! Check again.

6. I knew it could be solved by integrating under the integral sign. I just didn't think of integrating under the integral sign twice (which makes sense because there are two parameters).

7. Originally Posted by luobo
IT IS CORRECT. It turns out to be so simple that it may have surprised you! Check again.
it's not correct! you check again! haha ... the correct one is this: $\displaystyle \frac{\partial^2 F}{\partial a \; \partial b}\;=\;\int_0^\infty \frac{x^2}{(a^2x^2+1)(b^2x^2+1)} \; dx.$

8. Originally Posted by NonCommAlg
it's not correct! you check again! haha ... the correct one is this: $\displaystyle \frac{\partial^2 F}{\partial a \; \partial b}\;=\;\int_0^\infty \frac{x^2}{(a^2x^2+1)(b^2x^2+1)} \; dx.$
You are talking about the integration procedure, rather than the method itself. Again, the result turns out to be very simple or even simpler in terms of mathematical explanation. The method is correct and simpler anyway (at least in my view, Hahahaha...). Different from you, I like this method better. You may have a different view, that's fine.

I finally got some time to put other a detailed solution and here is the entire procedure:
$\displaystyle F(a,b)\;=\;\int_0^\infty \cot^{-1}(ax) \cot^{-1}(bx)\; dx$

$\displaystyle \frac{\partial^2 F}{\partial a \; \partial b}\;=\;\int_0^\infty \frac{x^2}{(a^2x^2+1)(b^2x^2+1)} \; dx\;=\frac{\pi}{2}\; \frac{1}{ab(a+b)}$

$\displaystyle \frac{\partial F}{\partial a}\;=\; \frac{\pi}{2a} \int_\infty^b \frac{1}{b(a+b)}\;db \;=\; \frac{\pi}{2a^2}\ln\frac{a}{a+b}$

$\displaystyle F(a,b)\;=\; \; \frac{\pi}{2} \int_\infty^a \frac{1}{a^2}\ln\frac{a}{a+b}\, da\;=\;\frac{\pi}{2}\left[\frac{1}{a} \ln \left(\frac{a+b}{b} \right) + \frac{1}{b} \ln \left(\frac{a+b}{a} \right) \right]$

We integrate from $\displaystyle \infty$ because when the integral is 0 when either $\displaystyle a\text{ or } b \text{ is } \infty$

9. Originally Posted by Random Variable
I knew it could be solved by integrating under the integral sign. I just didn't think of integrating under the integral sign twice (which makes sense because there are two parameters).
It happens sometimes. Some times you can even artificially introduce a variable to solve an integral. I bet you know that

10. It's not my turn, but no new integral has been posted. So I'm going to throw one out there.

$\displaystyle \int^{\infty}_{0} \frac{e^{-ax}-e^{-bx}}{x} \ dx \ \ (a, b >0)$

11. That $\displaystyle +$ is actually a $\displaystyle -.$ The solution is quite easy by noting that the integrand equals $\displaystyle \int_a^be^{-xt}\,dt.$

(I see you just edited the problem.)

12. Originally Posted by Random Variable
It's not my turn, but no new integral has been posted. So I'm going to throw one out there.

$\displaystyle \int^{\infty}_{0} \frac{e^{-ax}-e^{-bx}}{x} \ dx \ \ (a, b >0)$
Is this method OK?

we know $\displaystyle \int_0^{\infty} e^{-st} t^{n}~dt = \frac{\Gamma(n+1)}{s^{n+1}}$

so the integral equals to $\displaystyle \frac{ \Gamma(n) }{ a^n} - \frac{\Gamma(n)}{ b^n}$ $\displaystyle n \to 0$

$\displaystyle = \lim_{n\to 0} \frac{ b^n - a^n}{n}$

13. NonCommAlg :

I am very sure you have a lot of nice integrals , but not sure if you wanna share

14. Try this one:

Evaluate $\displaystyle \lim_{n\to\infty}\int_0^{\frac\pi3}\frac{\sin^nx}{ \sin^nx+\cos^nx}\,dx.$

15. Originally Posted by simplependulum
Is this method OK?

we know $\displaystyle \int_0^{\infty} e^{-st} t^{n}~dt = \frac{\Gamma(n+1)}{s^{n+1}}$

so the integral equals to $\displaystyle \frac{ \Gamma(n) }{ a^n} - \frac{\Gamma(n)}{ b^n}$ $\displaystyle n \to 0$

$\displaystyle = \lim_{n\to 0} \frac{ b^n - a^n}{n}$
I did it the way Krizalid suggested.

But I guess your way is correct too since $\displaystyle \lim_{n \to 0} \frac{b^{n}-a^{n}} {n} = \ln(b) - \ln(a) = \ln(b/a)$

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