Error in evaluating $\int_0^{\infty}\frac{\ln^2\left(x\right)}{\left(x^2+1\right)\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}dx$

Question

Quick disclaimer, I have never taken a calculus course so this is all based off of very informal understanding. This is the target integral $$ I=\int_0^{\infty}\frac{\ln^2\left(x\right)}{\left(x^2+1\right)\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}dx $$ I thought I could use some form of Feynman's technique so I started with this integral
$$ I\left(t\right)=\int_0^{\infty}\frac{x^t}{\left(x^2+1\right)\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)}dx $$ Differentiating twice would introduce the $\ln^2\left(x\right)$ I needed $$ I=I\ ''\left(0\right) $$

$$ I\left(t\right)=\int_0^{\infty}x^t\left(\frac{\left(Ax+B\right)}{x^2+1}+\frac{\left(Cx+D\right)}{x^2+4}+\frac{\left(Ex+F\right)}{x^2+9}+\frac{\left(Gx+H\right)}{x^2+16}\right)dx $$

$$ I\left(t\right)=\int_0^{\infty}x^t\left(\frac{\frac{1}{360}}{x^2+1}-\frac{\frac{1}{180}}{x^2+4}+\frac{\frac{1}{280}}{x^2+9}-\frac{\frac{1}{1260}}{x^2+16}\right)dx $$ After a partial fraction decomposition i had 4 similar integrals so i made a generalized function L using a as an input $$ L\left(a\right)=\int_0^{\infty}\frac{x^t}{x^2+a^2}dx=a^{t-1}\int_0^{\infty}\frac{x^t}{x^2+1}dx $$

$$ I\left(t\right)=\frac{L\left(1\right)}{360}-\frac{L\left(2\right)}{180}+\frac{L\left(3\right)}{280}-\frac{L\left(4\right)}{1260} $$ This is where i get a little unsure since i dont know if some of this is allowed but i do partial fraction and solve for A by letting x=i. I then equate A+B to $0$ so assuming t$\ne$1, A=-B $$ L\left(a\right)=a^{t-1}\int_0^{\infty}\left(\frac{A}{x-i}+\frac{B}{x+i}\right)dx$$

$$ \left(x+i\right)A+\left(x-i\right)B=x^t $$ $$ 2iA=i^t\ \ \Longrightarrow\ A=\frac{i^t}{2i}\ \Longrightarrow\ \ B=-\frac{i^t}{2i} $$

$$ L\left(a\right)=a^{t-1}\cdot\frac{i^t}{2i}\int_0^{\infty}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)dx $$ I then recombine the fractions and evaluate the integral $$ L\left(a\right)=a^{t-1}\cdot i^t\int_0^{\infty}\frac{1}{x^2+1}dx=a^{t-1}\cdot i^t\left(\arctan\left(\infty\right)-\arctan\left(0\right)\right) $$

$$ L\left(a\right)=\frac{\pi}{2}a^{t-1}i^t $$ I plug L back into my function I(t) then differentiate twice $$ I\left(t\right)=\frac{\pi i^t}{720}-\frac{\pi\left(2i\right)^t}{720}+\frac{\pi\left(3i\right)^t}{1680}-\frac{\pi\left(4i\right)^t}{10080}=\frac{\pi}{10080}\left(14i^t-14\left(2i\right)^t+6\left(3i\right)^t-\left(4i\right)^t\right) $$

$$ I'\left(t\right)=\frac{\pi}{10080}\left(14i^t\ln\left(i\right)-14\left(2i\right)^t\ln\left(2i\right)+6\left(3i\right)^t\ln\left(3i\right)-\left(4i\right)^t\ln\left(4i\right)\right) $$

$$ I''\left(t\right)=\frac{\pi}{10080}\left(14i^t\ln^2\left(i\right)-14\left(2i\right)^t\ln^2\left(2i\right)+6\left(3i\right)^t\ln^2\left(3i\right)-\left(4i\right)^t\ln^2\left(4i\right)\right) $$ We can now evaluate at zero $$ I''\left(0\right)=\frac{\pi}{10080}\left(14\ln^2\left(i\right)-14\ln^2\left(2i\right)+6\ln^2\left(3i\right)-\ln^2\left(4i\right)\right) $$

$$ \ln^2\left(ki\right)=\ln^2\left(i\right)+2\ln\left(k\right)\ln\left(i\right)+\ln^2\left(k\right)=-\frac{\pi^2}{4}+i\pi\ln\left(k\right)+\ln^2\left(k\right) $$

$$ I''\left(0\right)=\frac{\pi\left(4\left(-\frac{\pi^2}{4}+i\pi\right)-14\left(-\frac{\pi^2}{4}+i\pi\ln\left(2\right)+\ln^2\left(2\right)\right)+6\left(-\frac{\pi^2}{4}+i\pi\ln\left(3\right)+\ln^2\left(3\right)\right)-\left(-\frac{\pi^2}{4}+i\pi\ln\left(4\right)+\ln^2\left(4\right)\right)\right)}{10080} $$

$$ I''\left(0\right)=\frac{56i\pi^2-64i\pi^2\ln\left(2\right)-72\pi\ln^2\left(2\right)-5\pi^3+24i\pi^2\ln\left(3\right)+24\pi\ln^2\left(3\right)}{10080\cdot4} $$

This answer is not right. Where did I go wrong?

Answer 1

Too long for a comment.

In comments, @J.G. explained the problem.

But I think that using twice Feynman's trick makes, in this case, the problem more complicated than it is.

Using, as you did, partial fraction decomposition, you face a series of integrals $$I(a)= \int \frac{\log ^2(x)}{x^2+a^2}\,dx=\frac{i}{2 a}\Bigg(\int\frac{\log ^2(x)}{x+i a}\,dx-\frac{\log ^2(x)}{x-i a}\,dx\Bigg) $$ Consider $$J=\int\frac{\log ^2(x)}{x+b}\,dx$$ Perform two integrations by parts $$J=\log ^2(x) \log\left(1+\frac{x}{b}\right)+2 \log (x)\, \text{Li}_2\left(-\frac{x}{b}\right)-2 \text{Li}_3\left(-\frac{x}{b}\right)$$ Recombine the terms, use the bounds and simplify the complex terms to get $$K(a)= \int_0^\infty \frac{\log ^2(x)}{x^2+a^2}\,dx=\frac{\pi }{8 a} \left(4 \log ^2(a)+\pi^2\right)$$

Answer 2

Remarkably, the only result you really need is:

$$ R:= \int_0^{\infty} \frac{\ln^2{x}}{x^2+1} \, \mathrm {dx} = \frac{\pi^3}{8}$$

As you have observed, using partial fractions:

$$\begin{aligned} & ~~~~ \frac{1}{\left(x^2+1\right)\left(x^2+4\right)\left(x^2+9\right)\left(x^2+16\right)} \\& = \frac{1}{360} \cdot \frac{1}{1 + x^2} \, - \frac{1}{180}\cdot \frac{1}{4 + x^2} \, \\& + \frac{1}{280} \cdot \frac{1}{9 + x^2} -\frac{1}{1260} \cdot \frac{1}{16 + x^2} \, \end{aligned} $$

Finding the partial fractions, we find the integral is $I-J+K-L$ where:

$$ I := \frac{1}{360}\int_0^{\infty} \frac{\ln^2{x}}{1 + x^2} \, \mathrm{d}x$$ $$J := \frac{1}{180}\int_0^{\infty} \frac{\ln^2{x}}{4 + x^2} \, \mathrm{d}x$$ $$K := \frac{1}{280}\int_0^{\infty} \frac{\ln^2{x}}{9 + x^2} \, \mathrm{d}x$$ $$ L := \frac{1}{1260}\int_0^{\infty} \frac{\ln^2{x}}{16 + x^2} \, \mathrm{d}x$$

We only need to find one of these integrals; in fact, it suffices to prove the follwoing result:

$$ R:= \int_0^{\infty} \frac{\ln^2{x}}{x^2+1} \, \mathrm {dx} = \frac{\pi^3}{8}$$

I like the look of the $\displaystyle \frac{1}{1+x^2}$ because it reminds me of the classic geometric series, which converges uniformly over $(0,1)$. To exploit this, we write:

$$ R = \int_0^{1} \frac{\ln^2{x}}{x^2+1} \, \mathrm {dx} + \int_1^{\infty} \frac{\ln^2{x}}{x^2+1} \, \mathrm {dx}$$

Letting $x \mapsto \frac{1}{x}$ for the second integral we get:

$$ R = \int_0^{1} \frac{\ln^2{x}}{x^2+1} \, \mathrm {dx} + \int_0^{1} \frac{\ln^2{x}}{x^2+1} \, \mathrm {dx} = 2\int_0^{1} \frac{\ln^2{x}}{x^2+1} \, \mathrm {dx}$$

Now using the geometric series $\displaystyle \frac{1}{1+x^2} = \sum_{k \ge 0}(-1)^k x^{2k}$, we have

$$\displaystyle R = 2\int_0^{1} \frac{\ln^2{x}}{x^2+1} \, \mathrm {dx} = 2\int_0^{1} {\ln^2{x}} \sum_{k \ge 0}(-1)^k x^{2k} \, \mathrm {dx}$$

And switching the order of sum and integral, we get

$$R = 2\sum_{k \ge 0}(-1)^k \int_0^{1} {x^{2k}\ln^2{x}}\, \mathrm {dx}$$

Now consider $$f(k) = \int_0^{1} x^{2k}\, \mathrm {dx} = \frac{1}{2k+1}$$

Then taking the second derivative of both sides

$$f''(k) = 4\int_0^{1} x^{2k} \ln^2{x} \, \mathrm {dx} = \frac{8}{(2k+1)^3}$$

Therefore our integral is equal to the following known result:

$$\displaystyle R = 4 \sum_{k \ge 0} \frac{(-1)^k}{(2k+1)^3} = \frac{\pi^3}{8}$$.

It is very easy to calculate rest of the integrals in terms of $R$.

$$ J := \frac{1}{180}\int_0^{\infty} \frac{\ln^2{x}}{4 + x^2} \, \mathrm{d}x$$

Let $x \mapsto 2x$ then you get:

$$ \begin{aligned} J & = \frac{2}{180} \int_0^{\infty} \frac{\ln^2{2x}}{4+4x^2} \, \mathrm{d}x \\& = \frac{1}{360} \int_0^{\infty} \frac{\ln^2{2}}{1+x^2} + \frac{\ln^2{x}}{1+x^2} \, \mathrm{d}x \\& = \frac{\pi}{720} {\ln^2{2} }+\frac{1}{360}R \\& = \frac{\pi}{720} {\ln^2{2} }+\frac{\pi^3}{2880}\end{aligned} $$

Let $x \mapsto 4x$ then you get:

$$ \begin{aligned} L & = \frac{4}{1260} \int_0^{\infty} \frac{\ln^2{4x}}{16+16x^2} \, \mathrm{d}x \\& = \frac{1}{5040} \int_0^{\infty} \frac{\ln^2{4}}{1+x^2} + \frac{\ln^2{x}}{1+x^2} \, \mathrm{d}x \\& = \frac{\pi}{10080} {\ln^2{4} }+\frac{1}{5040}R \\& = \frac{\pi}{10080} {\ln^2{4} }+\frac{\pi^3}{40320}\end{aligned} $$

Let $x \mapsto 3x$ then you get:

$$ \begin{aligned} K & = \frac{3}{280} \int_0^{\infty} \frac{\ln^2{3x}}{9+9x^2} \, \mathrm{d}x \\& = \frac{1}{840} \int_0^{\infty} \frac{\ln^2{3}}{1+x^2} + \frac{\ln^2{x}}{1+x^2} \, \mathrm{d}x \\& = \frac{\pi}{1680} {\ln^2{3} }+\frac{1}{840}R \\& = \frac{\pi}{1680} {\ln^2{3} }+\frac{\pi^3}{6720}\end{aligned} $$

And finally we know $\displaystyle I = \frac{1}{360} R = \boxed{\frac{\pi^3}{2880}}$.

Your integral therefore evaluates to:

$$\begin{aligned} I-J+K-L & = \frac{\pi^3}{2880}-\left(\frac{\pi}{720} {\ln^2{2} }+\frac{\pi^3}{2880}\right)+\frac{\pi}{1680} {\ln^2{3} }\\&+\frac{\pi^3}{6720}-\left(\frac{\pi}{10080} {\ln^2{4} }+\frac{\pi^3}{40320}\right) \\& = \frac{1}{8064} \pi^3 - \frac{1}{560} \pi \ln^2 2 + \frac{1}{1680} \pi \ln^2 3. \end{aligned} $$