# Thread: Is this integral correct?

1. ## 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 $\int_{0}^{\infty} \frac{\ln x}{a^{2}+x^{2}} \ dx = \frac{\ln a}{a} \Big(\frac{\pi}{2}\Big) -\frac{1}{a} \Big(0\Big) = \frac{\pi \ln a}{2a}$ ?

2. Its correct according to Mathcad

3. Originally Posted by Calculus26
Thanks. It seemed to be correct for at least a few different values of a.

4. Mathcad does the symbolic calculation for arbitrary a-- its ruining my skills

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

6. 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.

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

8. nothing to see here

9. 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.