Finding Cauchy Principal Value for $\int_0^\infty \frac{\ln^2x }{(1-x)^2\sqrt{x}(x-4)} dx$

Question

I am trying to find Cauchy Principal value for $$\int_0^\infty \frac{\ln^2x }{(1-x)^2\sqrt{x}(x-4)} dx$$

Can you please suggest me where to start? Any help would be appreciated.

Thanks!

Answer 1

This can be done using complex analysis, but I am warning you that it is a pretty involved affair, deceptively so. After struggling with the problem of how to present the solution, I am going to lay out a systematic approach that ignores the motivation behind the steps. Think of it as a tradeoff between clarity and pedagogy if that makes any sense. I am favoring clarity.

So let's first get rid of that annoying square root by subbing $x \mapsto x^2$. The integral is then equal to

$$8 PV \int_0^{\infty} dx \frac{\log^2{x}}{(x^2-1)^2 (x^2-4)} $$

We note here that, while the pole at $x=1$ is indeed removable, the pole at $x=4$ is not. Thus the best we can do is compute this Cauchy principal value of the integral, which is the task at hand.

So let's consider the contour integrals

$$\oint_C dz \frac{\log^{k}{z}}{(z^2-1)^2(z^2-4)} $$

where $k \in \{1,2,3 \}$ and $C$ is the following keyhole contour of outer radius $R$ and inner radius $\epsilon$, with semicircular detours of radius $\epsilon$ about the poles at $z=1$ and $z=2$:

It will become apparent as we go why we need to consider three contour integrals rather than just one.

We now evaluate the contour integral by writing out the integral across each segment. Then the contour integral is equal to

$$\int_{\epsilon}^{1-\epsilon} dx \frac{\log^k{x}}{(x^2-1)^2 (x^2-4)} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^k{\left (1+\epsilon e^{i \phi} \right )}}{\left [\left (1+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (1+\epsilon e^{i \phi} \right )^2-4 \right ]} \\ + \int_{1+\epsilon}^{2-\epsilon} dx \frac{\log^k{x}}{(x^2-1)^2 (x^2-4)} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^k{\left (2+\epsilon e^{i \phi} \right )}}{\left [\left (2+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (2+\epsilon e^{i \phi} \right )^2-4 \right ]} \\ + \int_{2+\epsilon}^{R} dx \frac{\log^k{x}}{(x^2-1)^2 (x^2-4)} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{\log^k{\left (R e^{i \theta} \right )}}{\left [\left (R e^{i \theta} \right )^2-1 \right ]^2 \left [\left (R e^{i \theta} \right )^2-4 \right ]} \\ + \int_{R}^{2+\epsilon} dx \frac{(\log{x}+i 2 \pi)^k}{(x^2-1)^2 (x^2-4)}+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\left [\log{\left (2+\epsilon e^{i \phi} \right )} + i 2 \pi\right ]^k}{\left [\left (2+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (2+\epsilon e^{i \phi} \right )^2-4 \right ]} \\ + \int_{2-\epsilon}^{1+\epsilon} dx \frac{(\log{x}+i 2 \pi)^k}{(x^2-1)^2 (x^2-4)}+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\left [\log{\left (1+\epsilon e^{i \phi} \right )} + i 2 \pi\right ]^k}{\left [\left (1+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (1+\epsilon e^{i \phi} \right )^2-4 \right ]} \\ + \int_{1-\epsilon}^{\epsilon} dx \frac{(\log{x}+i 2 \pi)^k}{(x^2-1)^2 (x^2-4)}+ i \epsilon \int_{2 \pi}^{0} d\phi \, e^{i \phi} \frac{\left [\log{\left (\epsilon e^{i \phi} \right )} + i 2 \pi\right ]^k}{\left [\left (\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (\epsilon e^{i \phi} \right )^2-4 \right ]}$$

As $R \to \infty$, the sixth integral vanishes. As $\epsilon \to 0$, the twelfth integral vanishes. Further, the six odd-numbered integrals all combine to form the principal value integral

$$PV \int_0^{\infty} dx \frac{\log^k{x} - (\log{x} + i 2 \pi)^k}{(x^2-1)^2(x^2-4)} $$

We will now find the principal value integrals by going through the cases $k=1$, $k=2$, and $k=3$ in that order.

$k=1$

We evaluate the second, fourth, eighth, and tenth integrals above. In this case, then, the contour integral is equal to

$$-i 2 \pi PV \int_0^{\infty} \frac{dx}{(x^2-1)^2(x^2-4)} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log{\left (1+\epsilon e^{i \phi} \right )}}{\left [\left (1+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (1+\epsilon e^{i \phi} \right )^2-4 \right ]} \\ + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log{\left (2+\epsilon e^{i \phi} \right )}}{\left [\left (2+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (2+\epsilon e^{i \phi} \right )^2-4 \right ]} \\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\log{\left (2+\epsilon e^{i \phi} \right )} + i 2 \pi}{\left [\left (2+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (2+\epsilon e^{i \phi} \right )^2-4 \right ]} \\ + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\log{\left (1+\epsilon e^{i \phi} \right )} + i 2 \pi}{\left [\left (1+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (1+\epsilon e^{i \phi} \right )^2-4 \right ]}$$

Before we begin evaluation, it should be noted that the $PV$ is used in this integral loosely. As stated, even the Cauchy principal value of this integral does not converge in the limit as $\epsilon \to 0$. However, we will consider the limit not taken totally yet, but almost. (I know I am not being totally rigorous here, but I hope that I am being understood.) We will find that the $PV$ diverges in this limit, but we will also find that the divergent terms will eventual cancel in terms of the integrals we actually do need.

For what it's worth, denote

$$I_0 = PV \int_0^{\infty} \frac{dx}{(x^2-1)^2(x^2-4)} $$

Further, by the residue theorem, the contour integral is equal to $i 2 \pi$ times the sum of the residues at the poles. For the case $k=1$, denote the sum of the residues as $R_1$, i.e., it will be better if we put off computation of these until the end. Then, evaluating each of the above four integrals for small $\epsilon$ and rearranging, we find an expression for $I_0$:

$$-i 2 \pi I_0 + i \frac{\pi}{12} -i \frac{\pi \log{2}}{36} + \left (\frac{\pi^2}{18}-i \frac{\pi \log{2}}{36} \right ) + \left (-i \frac{\pi}{3 \epsilon} + \frac{\pi^2}{18}+i \frac{\pi}{12} \right ) = i 2 \pi R_1 $$

or

$$I_0 = -\frac1{6 \epsilon} + \frac1{12}-\frac{\log{2}}{36} - i \frac{\pi}{18} - R_1 $$

Again, note the term proportional to $1/\epsilon$, which is a consequence of the non convergence of this integral. It is OK to leave this as is because it will cancel other divergences later.

$k=2$

Using the notation established in the previous case, we have as the contour integral

$$-i 4 \pi I_1 +4 \pi^2 I_0 + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (1+\epsilon e^{i \phi} \right )}}{\left [\left (1+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (1+\epsilon e^{i \phi} \right )^2-4 \right ]} \\ + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (2+\epsilon e^{i \phi} \right )}}{\left [\left (2+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (2+\epsilon e^{i \phi} \right )^2-4 \right ]} \\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\left [\log{\left (2+\epsilon e^{i \phi} \right )} + i 2 \pi \right ]^2}{\left [\left (2+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (2+\epsilon e^{i \phi} \right )^2-4 \right ]} \\ + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\left [ \log{\left (1+\epsilon e^{i \phi} \right )} + i 2 \pi \right ]^2}{\left [\left (1+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (1+\epsilon e^{i \phi} \right )^2-4 \right ]}$$

where

$$I_1 = PV \int_0^{\infty} dx \frac{\log{x}}{(x^2-1)^2(x^2-4)} $$

In this case, $I_1$ is certainly finite. Then we expect the divergence in $I_0$ to cancel here. It will, as we will see. Invoking the residue theorem as we did above, with the sum of the residues of the $k=2$ contour integral being denoted as $R_2$, we have

$$-i 4 \pi I_1 + 4 \pi^2 \left ( -\frac1{6 \epsilon} + \frac1{12}-\frac{\log{2}}{36} - i \frac{\pi}{18} - R_1\right ) + 0 - i \frac{\pi \log^2{2}}{36} + \\ \frac{\pi^2\log{2}}{9} + i \left (\frac{\pi^2}{9} - \frac{\pi \log^2{2}}{36} \right ) + \left (\frac{2 \pi^2}{3 \epsilon} - \frac{\pi^2}{3} + i \frac{\pi^3}{9} \right ) = i 2 \pi R_2$$

or

$$I_1 = -\frac{\log^2{2}}{72} + i \pi R_1 - \frac12 R_2 $$

Note that, as expected, the divergences canceled and we are left with a nifty expression for the integral. We will do so for this last case, in which we get the integral that we finally care about.

$k=3$

The contour integral is now equal to

$$-i 6 \pi I_2 + 12 \pi^2 I_1 +i 8 \pi^3 I_0 + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^3{\left (1+\epsilon e^{i \phi} \right )}}{\left [\left (1+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (1+\epsilon e^{i \phi} \right )^2-4 \right ]} \\ + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^3{\left (2+\epsilon e^{i \phi} \right )}}{\left [\left (2+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (2+\epsilon e^{i \phi} \right )^2-4 \right ]} \\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\left [\log{\left (2+\epsilon e^{i \phi} \right )} + i 2 \pi \right ]^3}{\left [\left (2+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (2+\epsilon e^{i \phi} \right )^2-4 \right ]} \\ + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\left [ \log{\left (1+\epsilon e^{i \phi} \right )} + i 2 \pi \right ]^3}{\left [\left (1+\epsilon e^{i \phi} \right )^2-1 \right ]^2 \left [\left (1+\epsilon e^{i \phi} \right )^2-4 \right ]}$$

where

$$I_2 = PV \int_0^{\infty} dx \frac{\log^2{x}}{(x^2-1)^2(x^2-4)} $$

At this point, the reader should be able to figure out how to evaluate the integrals and substitutions and apply the residue theorem so I can just state the result:

$$I_2 = \frac{\pi^2 \log{2}}{54} - \frac{\pi^2}{18} - \frac{\log^3{2}}{108} + \frac{2 \pi^2}{3} R_1 + i \pi R_2 - \frac13 R_3 $$

where $R_3 $ is the sum of the residues of the contour integral for $k=3$.

Residue Computation

Finally, we have reduced the computation to the evaluation of the residues. We can combine the residue terms as we are effectively summing the residues of a single function. Thus

$$\frac{2 \pi^2}{3} R_1 + i \pi R_2 - \frac13 R_3 = \left [ \frac{d}{dz} \frac{\frac{2 \pi^2}{3} \log{z} + i \pi \log^2{z} - \frac13 \log^3{z}}{(z-1)^2 (z^2-4)} \right ]_{z=e^{i \pi}} + \frac{\frac{2 \pi^2}{3} (\log{2}+i \pi) + i \pi (\log{2}+i \pi)^2 - \frac13 (\log{2}+i \pi)^3}{(-2-1)^2(-2-2)}$$

The first term is the residue at the double pole $z=-1$, expressed so that $\arg{z} \in [0,2 \pi)$. The second term is the residue at $z=-2$. I leave the derivative computation for the reader or a CAS if the reader is so inclined. The result is

$$\frac{2 \pi^2}{3} R_1 + i \pi R_2 - \frac13 R_3 = -\frac{\pi^2}{36} + \frac{\log{2}}{108} \left (\pi^2 + \log^2{2} \right )$$

The final result

Combining the previous two results, we get an expression for $I_2$:

$$I_2 = \frac{\pi^2 \log{2}}{54} - \frac{\pi^2}{18} - \frac{\log^3{2}}{108} -\frac{\pi^2}{36} + \frac{\log{2}}{108} \left (\pi^2 + \log^2{2} \right )$$

and we may perform the cancelations. However, recall that the integral we want is $8$ times $I_2$. Thus, we finally write

$$PV \int_0^{\infty}dx \frac{\log^2{x}}{\sqrt{x} (1-x)^2 (x-4)} = \frac{2 \pi^2}{9} (-3 + \log{2})$$

which has been demonstrated by Cody and agrees with Mathematica.

Answer 2

Because I would rather not usurp the regular and very ingenious posters, I usually do not post many solutions. But, I hope no one minds I respond this time. Had some time before going to eat Christmas dinner :).

Ron, I would very much like to see your contour method!.

But, here is an approach to the problem. No doubt, someone has a more efficient method. When I see ln, often it signals DUTIS to introduce the ln term.

DUTIS stands for Differentiation Under The Integral Sign.

We'll differentiate twice, w.r.t 'a', then let $a\to -1/2$ in order to obtain the integrand we need.

Write the integrand as:

$$\int_{0}^{\infty}\frac{x^{a}}{(1-x)^{2}(x-4)}dx$$

$$=1/9\underbrace{\int_{0}^{\infty}\frac{x^{a}}{1-x}dx}_{\text{[1]}}-1/3\underbrace{\int_{0}^{\infty}\frac{x^{a}}{(1-x)^{2}}dx}_{\text{[2]}}+1/9\underbrace{\int_{0}^{\infty}\frac{x^{a}}{x-4}dx}_{\text{[3]}}$$

The first integral is a classic. It is related to digamma and evaluates to

[1]: $$\int_{0}^{\infty}\frac{x^{a}}{1-x}dx=\pi \cot(\pi a)$$

Diff twice w.r.t 'a', then let $a\to -1/2$:

$$\frac{d^{2}a}{da^{2}}\left[\frac{\pi}{9}\cot(\pi a)\right]=\frac{2}{9}\pi^{2}\cot(\pi a)\cdot \csc^{2}(\pi a)$$

$$\rightarrow \frac{2}{9}\pi^{2}\cot(\pi (-1/2))\cdot \csc^{2}(\pi (-1/2))=\boxed{0}$$

[2]: The middle integral can be done thusly:

$$1/3\int_{0}^{\infty}\frac{x^{a}}{(1-x)^{2}}dx=1/3\int_{0}^{1}\frac{x^{a}}{(1-x)^{2}}dx+1/3\int_{1}^{\infty}\frac{x^{a}}{(1-x)^{2}}dx$$

Let $x\to 1/x$ in the second integral. This leads to:

$$1/3\int_{0}^{1}\frac{x^{a}}{(1-x)^{2}}dx-1/3\int_{0}^{1}\frac{x^{-a}}{(1-x)^{2}}dx$$

Begin with $$\lim_{n\to \infty}\sum_{k=0}^{n}kx^{k-1\pm a}=\frac{x^{\pm a}}{(1-x)^{2}}$$

Integrate over $[0,1]$ and get sums related to digamma again:

$$\sum_{k=0}^{n}\left(\frac{k}{k-a}+\frac{k}{k+a}\right)=2(n+1)-a\left[\psi(a+n+1)-\psi(n-a+1)-\psi(a)+\psi(-a)\right]$$

I have to admit that I used tech to do the grunt work here.

Diff twice w.r.t 'a', then let $n\to \infty$.

This returns:

$$2\pi^{2}+2\pi^{2}\cot^{2}(\pi a)-2a\pi^{3}\cot(\pi a)-2a\pi^{3}\cot^{3}(\pi a)$$

Let $a\to -1/2$ and don't forget the -1/3 in front of the integral sign at the beginning.

$$\frac{1}{3}\left(2\pi^{2}+2\pi^{2}\cot^{2}(\pi (-1/2))-2(-1/2)\pi^{3}\cot(\pi (-1/2))-2(-1/2)\pi^{3}\cot^{3}(\pi (-1/2)) \right)= \boxed{\frac{2}{3}\pi^{2}}$$

[3]: Now we can do [3] by making a sub and using the digamma relation again.

$$1/9\int_{0}^{\infty}\frac{x^{a}}{x-4}dx$$

Make the sub $x=4y, \;\ dx=4dy$

This gives:

$$\frac{-4^{a}}{9}\int_{0}^{\infty}\frac{y^{a}}{1-y}dy$$

Now, we're back to the same integrand as in [1]. This then evaluates to:

$$\frac{-4^{a}}{9}\pi\cot(\pi a)$$

Diff this twice w.r.t. 'a':

$$\frac{2\cdot 4^{a}\pi}{9}\csc^{3}(\pi a)\left[2\cos(\pi a)\sin^{2}(\pi a)\ln^{2}(2)+\pi^{2}\cos(\pi a)-2\pi \ln(2)\sin(\pi a)\right]$$

Let $a\to -1/2$ and get $$\boxed{\frac{2\pi^{2}}{9}\ln(2)}$$

Put the 'boxed' pieces together and finally get the desired result of said integral:

$$\int_{0}^{\infty}\frac{\ln^{2}(x)}{(1-x)^{2}\sqrt{x}(x-4)}dx=[1]+[2]+[3]=\boxed{-\frac{2}{3}\pi^{2}+\frac{2\pi^{2}}{9}\ln(2)}$$

This appears to check numerically.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

• First, I'll evaluate $\ds{\left.\oint_{\cal C}{z^{\nu} \over z - a}\,\dd z\,\right\vert_{\substack{-1\ <\ \nu\ <\ 0\\[0.75mm] a\ >\ 0}}\,\,\,}$ where $\ds{\cal C}$ is a key-hole contour which "takes care" of the $\ds{z^{\nu}}$-branch cut $$ z^{\nu} = \verts{z}^{\nu}\expo{\ic\arg\pars{z}\nu}\,,\quad \arg\pars{z} \in \pars{0,2\pi}\,,\quad z \not= 0 $$

---

\begin{align} 0 & = \int_{0}^{\infty}{x^{\nu} \over x - a + \ic 0^{+}}\,\dd x + \int_{\infty}^{0} {x^{\nu}\expo{2\pi\nu\ic} \over x - a - \ic 0^{+}}\,\dd x \\[5mm] & = \mrm{P.V.}\int_{0}^{\infty}{x^{\nu} \over x - a}\,\dd x - \ic\pi a^{\nu} \\[2mm] & - \expo{2\pi\ic\nu}\,\mrm{P.V.} \int_{0}^{\infty}{x^{\nu} \over x - a}\,\dd x - \ic\pi a^{\nu}\expo{2\pi\ic\nu} \\[5mm] & = \pars{1 - \expo{2\pi\ic\nu}}\,\mrm{P.V.} \int_{0}^{\infty}{x^{\nu} \over x - a}\,\dd x - \ic\pi a^{\nu}\pars{1 + \expo{2\pi\ic\nu}} \end{align}

• Then \begin{align} \mrm{P.V.} \int_{0}^{\infty}{x^{\nu} \over x - a}\,\dd x & = \pi\ic a^{\nu}\,{1 + \expo{2\pi\ic\nu} \over 1 - \expo{2\pi\ic\nu}} = \pi\ic a^{\nu}\,{\expo{-\pi\ic\nu} + \expo{\pi\ic\nu} \over \expo{-\pi\ic\nu} - \expo{\pi\ic\nu}} \\[5mm] & = -\pi a^{\nu}\cot\pars{\pi\nu} \\[5mm] \mrm{P.V.} \int_{0}^{\infty}{\ln^{2}\pars{x} \over x - a} \,{\dd x \over \root{x}}& = \left.\partiald[2]{\bracks{-\pi a^{\nu}\cot\pars{\pi\nu}}} {\nu} \,\right\vert_{\,\nu\ =\ -1/2} = 2\pi^{2}\,{\ln\pars{a} \over \root{a}} \end{align}

• In addition, \begin{align} &\mrm{P.V.} \int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{x - a}\pars{x - 4}} \,{\dd x \over \root{x}} \\[5mm] = &\ {1 \over a - 4}\bracks{% \mrm{P.V.} \int_{0}^{\infty}{\ln^{2}\pars{x} \over x - a} \,{\dd x \over \root{x}} - \mrm{P.V.} \int_{0}^{\infty}{\ln^{2}\pars{x} \over x - 4} \,{\dd x \over \root{x}}} \\[5mm] = &\ 2\pi^{2}\,{\ln\pars{a}/\root{a} - \ln\pars{2} \over a - 4} \end{align}

• Finally, \begin{align} &\mrm{P.V.} \int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{x - 1}^{2}\pars{x - 4}}\,{\dd x \over \root{x}} \\[5mm] = &\ 2\pi^{2}\,\totald{}{a}\bracks{% {\ln\pars{a}/\root{a} - \ln\pars{2} \over a - 4}}_{\,a\ =\ 1} \\[5mm] = &\ \bbx{{2 \over 9}\,\pi^{2}\ln\pars{2} - {2 \over 3}\,\pi^{2}} \approx -5.0595 \\ & \end{align}