# Is this integral correct?

• August 19th 2009, 09:24 PM
Random Variable
Is this integral correct?
I've been working on this for the past hour.

$\int_{0}^{\infty} \frac{\ln x}{a^{2}+x^{2}} \ dx \ (a>0)$

let $u = \frac{a}{x}$

then $du = -\frac{a}{x^{2}}dx = -\frac{u^{2}}{a} dx$

$\int_{0}^{\infty} \frac{\ln x}{a^{2}+x^{2}} \ dx = -\int_{\infty}^{0} \frac{\ln\Big( \frac{a}{u}\Big)}{a^{2}+ \frac{a^{2}}{u^{2}}} \ \frac{a}{u^{2}} du =$ $\frac{1}{a} \int_{0}^{\infty} \frac{\ln\Big( \frac{a}{u}\Big)}{u^{2}+1} \ du$

$= \frac{\ln a}{a} \int_{0}^{\infty} \frac{1}{u^{2}+1} \ du -\frac{1}{a} \int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du$

$\int_{0}^{\infty} \frac{1}{u^{2}+1} \ du = \tan^{-1} (\infty) - \tan^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$

$\int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du$

let $w = \frac{1}{u}$

then $dw = -\frac{1}{u^{2}} du = - w^{2} du$

$\int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du = -\int_{\infty}^{0} \frac{\ln \Big( \frac{1}{w}\Big)}{\frac{1}{w^{2}}+1} \ \frac{1}{w^{2}} \ dw$ $= -\int_{0}^{\infty} \frac{\ln w}{w^{2}+1} \ dw$

which implies that $\int_{0}^{\infty} \frac{\ln u}{u^{2}+1} \ du =0$

so finally $\int_{0}^{\infty} \frac{\ln x}{a^{2}+x^{2}} \ dx = \frac{\ln a}{a} \Big(\frac{\pi}{2}\Big) -\frac{1}{a} \Big(0\Big) = \frac{\pi \ln a}{2a}$ ?
• August 19th 2009, 10:38 PM
Calculus26
• August 19th 2009, 10:51 PM
Random Variable
Quote:

Originally Posted by Calculus26

Thanks. It seemed to be correct for at least a few different values of a.
• August 19th 2009, 10:53 PM
Calculus26
Mathcad does the symbolic calculation for arbitrary a-- its ruining my skills
• August 20th 2009, 12:41 AM
Bruno J.
Do you know the residue theorem? I haven't checked fully but it seems that such an integral would yield fairly easily to it.
• August 20th 2009, 04:42 AM
Moo
Hello,

As Bruno J. mentioned, it can be done with the residue theorem.

If you consider the keyhole contour, then it can be proved that :

$\int_0^\infty R(x)\ln(x) ~dx=-\frac 12 ~ \Re \left(\sum \text{Res}\left\{R(z)(\log(z))^2\right\}\right)$

where the branch of the logarithm is such that $-\pi<\text{arg}(z)\leq \pi$

It's not more beautiful than your method though.
• August 20th 2009, 10:05 AM
Random Variable
Quote:

Originally Posted by Bruno J.
Do you know the residue theorem? I haven't checked fully but it seems that such an integral would yield fairly easily to it.

I didn't even think about using contour integration. The problem is actually an exercise in my complex analysis textbook.

Moo suggested using the keyhole contour. My book suggests using a semicircle in the upper-half plane with a little semicircle about the origin (because ln z is undefined for z=0). They also suggest using the branch $\log_{-\frac{\pi} {2}} = ln|z| + i \theta \ (-\frac{\pi}{2} < \theta \le \frac{3 \pi}{2})$ so that ln z is analytic on and inside of the contour.

So let R be the radius of the big semicircle and r be the radius of the small semicircle

and let $f(z) = \frac{\log_{\frac{-\pi}{2}} z}{a^{2}+z^{2}}$

then $\int_{C} f(z) dz = \int^{-r}_{-R} f(x) dx + \int_{-C_{r}} f(z) dz + \int^{R}_{r} f(x) dx + \int_{C_{R}} f(z) dz$

$= \int^{-r}_{-R} \frac{ \ln |x| + i \pi}{a^{2}+z^{2}} \ dx + \int_{-C_{r}} f(z) dz + \int^{R}_{r} \frac {\ln x}{a^{2} + x^{2}} \ dx + \int_{C_{R}} f(z) dz$

$\int_{C} f(z) dz = \int_{C} \frac{\log_{-\frac{\pi}{2}} z} {a^{2}+z^{2}} \ dz = \int_{C} \frac{\log_{-\frac{\pi}{2}} z} {(z-ai)(z+ai)} \ dz$

so there is a simple pole inside of the contour at $z_{0} = ai$

$\text{Res} \{f,ai\} = \lim_{z \to ai} \frac{\log_{-\frac{\pi}{2}} z}{z+ai} = \frac{\ln|ai| + i\frac{\pi}{2}}{2ai} = \frac{\ln a + i\frac{\pi}{2}}{2ai}$

then $\int_{C} f(z) dz = 2 \pi i \ \text{Res} \{f, ai\} = \frac{\pi \ln a}{a} + i \frac{\pi^{2}}{2a}$

So I have $\int^{-r}_{-R} \frac{ \ln |x| + i \pi}{a^{2}+z^{2}} \ dx + \int_{-C_{r}} f(z) dz + \int^{R}_{r} \frac {\ln x}{a^{2} + x^{2}} \ dx + \int_{C_{R}} f(z) dz = \frac{\pi \ln a}{a} + i \frac{\pi^{2}}{2a}$

now letting R go to infinity and r go to zero

$\int^{0}_{-\infty} \frac{ \ln |x| + i \pi}{a^{2}+z^{2}} \ dx + 0 + \int^{\infty}_{0} \frac {\ln x}{a^{2} + x^{2}} \ dx + 0 = \frac{\pi \ln a}{a} + i \frac{\pi^{2}}{2a}$

I'm just assuming that the two contour integrals evaluate to zero. I'll try to offer a justification in another post. But I'll probably need some help.

Equating real parts on both sides I have

$\int^{0}_{-\infty} \frac{ \ln |x|}{a^{2}+z^{2}} \ dx + \int^{\infty}_{0} \frac {\ln x}{a^{2} + x^{2}} \ dx = \int^{\infty}_{0} \frac{ \ln x }{a^{2}+z^{2}} \ dx + \int^{\infty}_{0} \frac {\ln x}{a^{2} + x^{2}} \ dx =$ $\frac{\pi \ln a}{a}$

so finally $\int^{\infty}_{0} \frac {\ln x}{a^{2} + x^{2}} \ dx = \frac{\pi \ln a}{2a}$

(Whew)
• August 20th 2009, 11:16 AM
Random Variable
nothing to see here
• August 20th 2009, 06:08 PM
Random Variable
OK. After a bit of confusion about the reverse triangle inequality, I think I can show that

$\lim_{R \to \infty} \int_{C_{R}} f(z) dz = \lim_{r \to 0}\int_{-C_{r}} f(z) dz = 0$

$\Big|\int_{C_{R}} f(z) dz \Big|= \Big|\int_{C_{R}} \frac{\log_{-\frac{\pi}{2}}z}{a^{2}+z^{2}} \ dz\Big|$

let $z = Re^{i \theta}$

then $dz = iRe^{i \theta} d \theta$

$= \Big|\int_{0}^{\pi}\frac{\ln|Re^{i \theta}| + i \theta}{a^{2} + (Re^{i \theta})^{2}} iRe^{i \theta} \ d \theta \Big|$ $= \int^{\pi}_{0} \Big|\frac{\ln R + i \theta}{a^{2} + (Re^{i \theta})^{2}}\Big| \Big|iRe^{i \theta}\Big| \ d \theta$

$\int^{\pi}_{0} \Big|\frac{\ln R + i \theta}{a^{2} + (Re^{i \theta})^{2}}\Big| R \ d \theta \le \int^{\pi}_{0} \frac{R \ln R + R\theta}{|a^{2} + (Re^{i \theta})^{2}|} \ d \theta \le \int^{\pi}_{0} \frac{R \ln R + R \theta}{-a^{2} + R^{2}} \ d \theta \ \ (\text{since} \ R >a)$

$= \frac{R \ln R}{-a^{2} + R^{2}} \int_{0}^{\pi} d \theta + \frac{R}{-a^{2}+R^{2}}\int_{0}^{\pi} \theta \ d \theta$

$= \frac{2\pi R\ln R + \pi^{2}}{-2(a^{2}+R^{2})}$

$\lim_{R \to \infty} \frac{2\pi R\ln R + \pi^{2}}{2(-a^{2}+R^{2})} = \lim_{R \to \infty} \frac{2 \pi \ln R + 2 \pi}{4R} = 0$

which according to the ML inequality shows that $\lim_{R \to \infty}\int_{C_{R}} f(z) dz = 0$

The other one is similar except r<a.