Integral ∫∞0arccot(√x−2√x+1)x+1dx\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx

Is it possible to evaluate this integral in a closed form?
0arccot(x2x+1)x+1dx

Answer

Define the function I(s) for s>0 by

I(s)=0arccot(xesx+1)x+1dx.

By observing that I()=0, we have

I(s)=sI(t)dt.

Applying Leibniz’s integral rule,

I(s)=0esx+11+(xesx+1)2dxx+1.

Now with the substitution x=tan2θ,

I(s)=π20sinθcoshssinθ=12π2π2cosθcoshscosθ.

Then with the substitution z=eiθ, it follows that

I(s)=i2Γz2+1z22zcoshs+1dzz,

where the contour Γ denotes the counter-clockwise semicircular arc joining from i to i. Note that we have z22zcoshs+1=0 if and only if coshs=cosθ if and only if z=eiθ=e±s. Thus deforming the contour to the straight line joining from i to i with infinitesimal indent at the origin, we obtain

I(s)=i2{PViif(z)dz+2πiRes(f,es)+πiRes(f,0)},

where f denotes the integrand

f(z)=z2+1z(z22zcoshs+1).

Proceeding the calculation,

I(s)=i2{iPV11f(ix)dx+πi2πicoths}=1210{f(ix)+f(ix)}dxπ2+πcoths=coshs1012(1x2){12(x+x1)}2+sinh2sdxπ2+πcoths.

Now with the substitution u=12{x+x1}, it follows that

I(s)=coshs1duu2+sinh2sπ2+πcoths=arctan(sinhs)cothsπ2+πcoths.

Finally, with the substitution x=sinht,

I(s)=sI(t)dt=s{arctan(sinht)cotht+π2πcotht}dt=sinhs(arctanxx+π2x2+1πx)dx.

Our problem corresponds to s=log2, or equivalently, a:=sinhs=34.

a(arctanxx+π2x2+1πx)dx=aarctan(1/x)xdx+π2a(1x2+11x)dx=1/a0arctanxxdx+π2lim

where the function

\operatorname{Ti}(x) = \frac{\operatorname{Li}_{2}(ix) – \operatorname{Li}_{2}(-ix)}{2i}

is the inverse tangent integral. Using the following simple identity

\operatorname{Ti}(x) = \operatorname{Ti}\left(\frac{1}{x}\right) + \frac{\pi}{2}\log x,

it follows that

\begin{align*}
\int_{a}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} – \frac{\pi}{x} \right) \, dx
= – \operatorname{Ti}(a) – \frac{\pi}{2} \log\left( \frac{a + \sqrt{a^{2} + 1}}{2a^{2}} \right).
\end{align*}

Therefore, plugging a = \frac{3}{4} gives

\int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} – 2\sqrt{x+1} )}{x+1} \, dx
= \pi \log(3/4) – \operatorname{Ti}(3/4)

as Cleo pointed out without explanation. More generally, for k > 1

\int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} – k \sqrt{x+1} )}{x+1} \, dx
= \frac{\pi}{2} \log \left( \frac{(k^{2} – 1)^{2}}{2k^{3}} \right) – \operatorname{Ti}\left( \frac{k^{2} – 1}{2k} \right).

Attribution
Source : Link , Question Author : Laila Podlesny , Answer Author : Sangchul Lee

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