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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?

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  • $\begingroup$ If you set $$q(u)=\int_0^\infty \frac{x^u}{(\ln^2 x+\pi^2/4)(1+x^2)^2}dx$$ Then the integral is given by $$\pi^3q(3)-6\pi^2q'(2)+12\pi q''(1)-8q'''(0)$$ $\endgroup$
    – clathratus
    Commented Jun 4, 2019 at 18:36
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    $\begingroup$ The function $q$, mentioned in my previous comment, can be written as $$q(s)=\frac{1}{4\pi}\int_0^{\infty}\frac{\cosh(\frac{\pi(s-1)}{2}t)\text{sech}^2(\frac{\pi}{2}t)}{t^2+1}dt$$ $\endgroup$
    – clathratus
    Commented Jun 5, 2019 at 4:27

1 Answer 1

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We can use some instances of Schroder's Integral, which evaluates Gregory coefficients, namely:

$$(-1)^{n-1}G_n=\int_0^\infty\frac{1}{(\pi^2+\ln^2 t)(1+t)^n}dt;\quad G_1=\frac12, \ G_2=-\frac{1}{12}$$


But let's adjust things first.

$$I=\int_0^\infty \frac{(\pi x - 2\log{x})^3}{\left(\log^2{x} + \frac{\pi^2}{4}\right)(1+x^2)^2} dx \overset{x^2=t}=2\int_0^\infty \frac{(\pi\sqrt t-\ln t)^3}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$

$$=2\pi^3\int_0^\infty \frac{t}{(\pi^2+\ln^2 t)(1+t)^2}dt-6\pi^2 \int_0^\infty \frac{\sqrt t\ln t}{(\pi^2+\ln^2 t)(1+t)^2}dt$$

$$+6\pi \int_0^\infty \frac{\ln^2 t}{(\pi^2+\ln^2 t)(1+t)^2}dt-2\int_0^\infty \frac{\ln^3 t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$

$$=2\pi^3 I_1-6\pi^2I_2+6\pi I_3-2 I_4$$


$$I_1=\int_0^\infty \frac{t}{(\pi^2+\ln^2 t)(1+t)^2}dt=\int_0^\infty \frac{(1+t)-1}{(\pi^2+\ln^2 t)(1+t)^2}dt$$

$$=\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)}dt-\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)^2}dt$$

$$=G_1+G_2=\frac12-\frac{1}{12}=\frac{5}{12}$$


The $I_2$ integral can be found here.

$$I_2=\int_0^\infty \frac{\sqrt t\ln t}{(\pi^2+\ln^2 t)(1+t)^2}dt\overset{t=\frac{1}{x}}=-\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2}\frac{dx}{\sqrt x}=\frac{\pi}{24}$$


$$I_3=\int_0^\infty \frac{\ln^2 t}{(\pi^2+\ln^2 t)(1+t)^2}dt = \int_0^\infty \frac{(\pi^2 +\ln^2 t)-\pi^2}{(\pi^2+\ln^2 t)(1+t)^2}dt$$

$$=\int_0^\infty \frac{1}{(1+t)^2}dt-\pi^2\int_0^\infty \frac{1}{(\pi^2+\ln^2 t)(1+t)^2}dt$$

$$=1+\pi^2 G_2=1-\frac{\pi^2}{12}$$


$$I_4=\int_0^\infty \frac{\ln^3 t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}=\int_0^\infty \frac{\ln t((\pi^2+\ln^2 t ) -\pi^2 )}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$

$$=\int_0^\infty \frac{\ln t}{(1+t)^2}\frac{dt}{\sqrt t}-\pi^2 \int_0^\infty \frac{\ln t}{(\pi^2+\ln^2 t)(1+t)^2}\frac{dt}{\sqrt t}$$

$$=-\pi+\pi^2 I_2=-\pi +\frac{\pi^3}{24}$$


$$\require{cancel} \Rightarrow I=\cancel{2\pi^3{\frac{5}{12}}} -\cancel{6\pi^2{\frac{\pi}{24}}}+6\pi \left({1-\cancel{\frac{\pi^2}{12}}}\right)-2\left({-\pi +\cancel{\frac{\pi^3}{24}}}\right)={8\pi}$$

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