# For $a>0$ compute $\int_0^\infty \frac{\log x}{(a + x)^3}\,dx$

Assume that $a > 0$. Compute $$\int_0^\infty \frac{\log x}{(a + x)^3}\,dx.$$

This seems resistant to the usual strategies—indeed, I can't even think of an appropriate contour.

I tried shifting the integral, $$\int_a^\infty \frac{\log (x - a)}{x^3}\,dx,$$ but this seemed to cause sections of my contour to be unbounded.

I couldn't find a substantially similar question here, but please let me know if it has already been asked.

• Have you integrate something with a log and a polynomial in the denominator before? Because there are standard tricks one uses to find such integrals. – Shashi Jun 21 '18 at 18:23
• I have, but the tricks that I've used don't seem to work here. For example, there is no $\theta$ such that $(a + xe^{\theta i})^3$ is a multiple of $(a + x)^3$. (I'm comfortable with denominators like $a+x^n$.) – Peter Kagey Jun 21 '18 at 18:29
• Never heard about something like considering the same integral, but with $log^2(x)$ instead of $\log(x)$? – Shashi Jun 21 '18 at 18:34
• Perhaps the feynman technique on $\displaystyle\int_0^\infty\frac{\ln(bx)}{(x+a)^3}\,dx$ could work – The Integrator Jun 21 '18 at 18:37
• @The Integrator Yeah the DUI method will work greatly but then the only problem is how are you going to evaluate the constant of integration afterwards – Darkrai Jun 21 '18 at 19:05

## 3 Answers

$\newcommand{\Log}{\operatorname{Log}}\newcommand{\Res}{\operatorname{Res}}$Hint on using Residue Calculus

Being a little bit stubborn, I will consider the following integral: \begin{align} \int_C \frac{\Log^2(-z)}{(z+a)^3}\,dz \end{align} where $C$ is the keyhole contour with the "key" on the positive axis and $\Log(\cdot)$ the pricipal value of the logarithm. This is the trick that I was talking about, I thought it was standard.

The circle has radius $R>a$. Let $C_R$ be the circular arc and $K^+_{R,\varepsilon}$ be the part of the "key" above the $x$-axis and $K^-_{R,\varepsilon}$ the one below. The $\varepsilon>0$ makes sure that there is some little space so that the arc on the positive real axis stays a little bit away from the axis (You should read why one does this). Of course after letting $R$ go to infinity and $\varepsilon\to 0$ the integral over the circle arc vanishes, together with the residue theorem, you end up with: \begin{align}\tag{1} \int^\infty_0 \frac{(\log(t)-i\pi)^2}{(t+a)^3} -\frac{(\log(t)+i\pi)^2}{(t+a)^3}\,dt= \Res_{z=a} \frac{\Log^2(-z)}{(z+a)^3} \end{align} Okay I did skip some steps so that you fill them in. However that form over there is to let you know that you will get to deal with two integrals over the positive real axis and each will give one part of the LHS of (1). After doing some simplifcation you will see the light, namely $\log(t)$ appears there and that is our original integral. Let me know if something won't work out.

Using \begin{eqnarray} &&\int\frac{\log x}{(a + x)^3}\,dx\\ &=&-\frac12\int \log xd(a + x)^{-2}\\ &=&-\frac12(x+a)^{-2}\log x+\frac12\int\frac{1}{x(x+a)^2}dx\\ &=&-\frac12(x+a)^{-2}\log x+\frac1{2a^2}\int\bigg[\frac1x-\frac1{x+a}-\frac{1}{(x+a)^2}\bigg]dx\\ &=&-\frac12(x+a)^{-2}\log x+\frac1{2a^2}\bigg[\log x-\log(x+a)+\frac{1}{x+a}\bigg]+C \end{eqnarray} it is easy to obtain $$\int_a^\infty \frac{x}{(x - a)^3}\,dx=\frac{1-\log a}{2a^2}.$$

• $log(0)$ does not exist so i think you should rethink your answer – The Integrator Jun 21 '18 at 18:39
• @TheIntegrator, use $\lim_{x\to0^+}$. – xpaul Jun 21 '18 at 18:42

It is not resistant to usual strategies. Assuming $a>0$ and setting $x=a z$ we have $$\int_{0}^{+\infty}\frac{\log x}{(a+x)^3}\,dx = \frac{1}{a^3}\int_{0}^{+\infty}\frac{\log a+\log z}{(z+1)^3}\,dz$$ which just depends on two simple integrals: $$\int_{0}^{+\infty}\frac{dz}{(z+1)^3}=\frac{1}{2},\qquad \int_{0}^{+\infty}\frac{\log z}{(z+1)^3}\,dz = -\frac{1}{2}.$$ The latter can be simply computed by integration by parts, reducing the integrand function to a rational function.