Prove $$\int_0^\infty \frac{(\pi x - 2\log{x})^3}{\left(\log^2{x} + \frac{\pi^2}{4}\right)(1+x^2)^2} \text{d}x = 8\pi$$
I tried using a modified version of the integrand and an origin-indented semicircular contour on the positive real half of the complex plane.
Despite my best efforts, I am unable to retrieve the desired integral, as the negative factor from the negative imaginary axis is preventing me from being able to simplify the integrals.
On top of that, the residue doesn't come anywhere close to the exact result.
Are there better methods that I am missing?
Answer
Before we start I should mention that we will use some instances of Schroder's Integral which evaluates Gregory coefficients, namely:
$$\sf (-1)^{n-1}G_n=\int_0^\infty\frac{1}{(\pi^2+\ln^2 t)(1+t)^n}dt$$
But let's adjust things first.
$$\sf \color{orange}{I=\int_0^\infty \frac{(\pi x - 2\log{x})^3}{\left(\log^2{x} + \frac{\pi^2}{4}\right)(1+x^2)^2} dx} \overset{x^2=t}=2\int_0^\infty \frac{(\pi\sqrt t-\ln t)^3}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$\sf =2\pi^3\int_0^\infty \frac{t}{(\pi^2+\ln^2 t)(1+t)^2}dt-6\pi^2 \int_0^\infty \frac{\sqrt t\ln t}{(\pi^2+\ln^2 t)(1+t)^2}dt$$
$$\sf +6\pi \int_0^\infty \frac{\ln^2 t}{(\pi^2+\ln^2 t)(1+t)^2}dt-2\int_0^\infty \frac{\ln^3 t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$
$$\sf =2\pi^3 I_1-6\pi^2I_2+6\pi I_3-2 I_4$$
Evaluation of $I_1$.
$$\tt \color{blue}{I_1=\int_0^\infty \frac{(1+t)-1}{(\pi^2+\ln^2 t)(1+t)^2}dt=\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)}dt-\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)^2}dt}$$
$$\tt \color{blue}{=G_1+G_2=\frac12-\frac{1}{12}=\frac{5}{12}}$$
Evaluation of $I_2$.
I have solved this integral here.
$$\tt \color{red}{I_2=\int_0^\infty \frac{\sqrt t\ln t}{(\pi^2+\ln^2 t)(1+t)^2}dt\overset{t=\frac{1}{x}}=-\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2}\frac{dx}{\sqrt x}=\frac{\pi}{24}}$$
Evaluation of $I_3$.
$$\tt \color{purple}{I_3= \int_0^\infty \frac{(\pi^2 +\ln^2 t)-\pi^2}{(\pi^2+\ln^2 t)(1+t)^2}dt=\int_0^\infty \frac{1}{(1+t)^2}dt-\pi^2\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)^2}dt}$$
$$\tt \color{purple}{=1+\pi^2 G_2=1-\frac{\pi^2}{12}}$$
Evaluation of $I_4$.
Write $\ln^3 t= \ln t((\pi^2+\ln^2 t ) -\pi^2 )$ to get:
$$\tt \color{green}{I_4=\int_0^\infty \frac{\ln^3 t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}=\int_0^\infty \frac{\ln t}{(1+t)^2}\frac{dt}{\sqrt t}-\pi^2 \int_0^\infty \frac{\ln t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}}$$
$$\tt \color{green}{=-\pi+\pi^2 I_2=-\pi +\frac{\pi^3}{24}}$$
$$\require{cancel} \sf \Rightarrow \color{orange}I=\cancel{2\pi^3\color{blue}{\frac{5}{12}}} -\cancel{6\pi^2\color{red}{\frac{\pi}{24}}}+6\pi \left(\color{purple}{1-\cancel{\frac{\pi^2}{12}}}\right)-2\left(\color{green}{-\pi +\cancel{\frac{\pi^3}{24}}}\right)=\color{orange}{8\pi}$$
We actually only used $G_1$ and $G_2$ and I'm sure that we can evaluate them using elementary methods, atleast the first one is pretty easy, but for the second one we might have to strive a little.
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