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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.
$\newcommand{\Log}{\operatorname{Log}}\newcommand{\Res}{\operatorname{Res}}$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 $$ \int_a^\infty \frac{x}{(x - a)^3}\,dx=\frac{1-\log a}{2a^2}.$$
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.
