1. ## improper integral #4

Evaluate $\displaystyle \int_{0}^{\infty} \frac{\ln x}{(x+a)(x+b)} \ dx \ \ a,b>0 \ (a \ne b)$

Unlike the last integral, try not to use contour integration.

2. ## Re: improper integral #4

Originally Posted by Random Variable
Evaluate $\displaystyle \int_{0}^{\infty} \frac{\ln x}{(x+a)(x+b)} \ dx \ \ a,b>0 \ (a \ne b)$

Unlike the last integral, try not to use contour integration.
Spoiler:

We first make the observation that

$\displaystyle \displaystyle \frac{1}{(x+a)(x+b)}=\frac{1}{b-a}\int_a^b\frac{dy}{(x+y)^2} \text{ }dy$

And thus, if $\displaystyle I(a,b)$ is the given integral we have that

$\displaystyle \displaystyle I(a,b)=\frac{1}{b-a}\int_0^\infty \int_a^b \frac{\log(x)}{(x+y)^2}\text{ }dy\,dx$

Now, since

$\displaystyle \displaystyle \int_{\mathbb{R}\times[a,b]}\left|\frac{\log(x)}{(x+y)^2}\right|\text{ }d(x,y)<\infty$

(which is easily verified via estimation) we may apply Fubini's theorem to conclude that

$\displaystyle \displaystyle I(a,b)=\frac{1}{b-a}\int_a^b\int_0^\infty \frac{\log(x)}{(x+y)^2}\text{ }dx\,dy$

But, it is easy to see that

$\displaystyle \displaystyle \int\frac{\log(x)}{(x+y)^2}\text{ }dx=\frac{x\log(x)}{y(y+x)}-\frac{\log(y+x)}{y}$

So that

$\displaystyle \displaystyle \int_0^{\infty}\frac{\log(x)}{(x+y)^2}\text{ }dx=\frac{\log(y)}{y}$

and so

\displaystyle \begin{aligned}I(a,b) &=\frac{1}{b-a}\int_a^b\int_0^\infty\frac{\log(x)}{(x+y)^2} \text{ }dx\\ &=\frac{1}{b-a}\int_a^b\frac{\log(y)}{y}\;dy\\ &=\frac{\log^2(b)-\log^2(a)}{2(b-a)}\end{aligned}

3. ## Re: improper integral #4

There is an easier approach.

Make the substitution $\displaystyle u = \frac{ab}{x}$ and see what happens.

4. ## Re: improper integral #4

Originally Posted by Random Variable
There is an easier approach.

Make the substitution $\displaystyle u = \frac{ab}{x}$ and see what happens.
Yes, that may work. But it wasn't what was most natural to me. I evaluated the integral on Wolfram Alpha for a couple of specific values and noticed that it was always of the form $\displaystyle F(b)-F(a)$ where $\displaystyle b>a$ that just screamed use a double integral to me.

5. ## Re: improper integral #4

Originally Posted by Drexel28
Yes, that may work. But it wasn't what was most natural to me. I evaluated the integral on Wolfram Alpha for a couple of specific values and noticed that it was always of the form $\displaystyle F(b)-F(a)$ where $\displaystyle b>a$ that just screamed use a double integral to me.
The substitution is not something I came up with. My automatic approach would be to find the Cauchy principal value using contour integration and then drop the P.V. label because the integral is convergent.

But if you make that substitution you get

$\displaystyle \int^{\infty}_{0} \frac{ \ln x}{(x+a)(x+b)} \ dx = \ln (ab) \int^{\infty}_{0} \frac{du}{(u+a)(u+b)} - \int^{\infty}_{0} \frac{ \ln u}{(u+a)(u+b)} \ du$

so $\displaystyle \int^{\infty}_{0} \frac{\ln x}{(x+a)(x+b)} \ dx = \frac{\ln (ab)}{2} \int^{\infty}_{0} \frac{du}{(x+a)(x+b)} \ dx$

EDIT: $\displaystyle = \frac{\ln a + \ln b}{2(a-b)} \int^{\infty}_{0} \Big(\frac{1}{x+b} - \frac{1}{x+a} \Big) \ dx = \frac{\ln a + \ln b}{2(a-b)} \ (\ln a - \ln b)$

$\displaystyle = \frac{\ln^{2} a - \ln^{2} b}{2(a-b)}$

6. ## Re: improper integral #4

The same substitution can be used to evaluate $\displaystyle \int^{b}_{a} \frac{\ln x}{(x+a)(x+b)} \ dx$.