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5

For first, we have: $$I=\Re\,\text{Li}_2(1\pm i\sqrt{3})=\Re\int_{1}^{i\sqrt{3}}\frac{\log t}{1-t}\,dt=\frac{\pi^2}{6}+\Re\int_{0}^{i\sqrt{3}}\frac{\log t}{1-t}\,dt\tag{1}$$ hence: $$I = \frac{\pi^2}{6}-\int_{0}^{\sqrt{3}}\frac{\frac{\pi}{2}+t\log t}{1+t^2}\,dt\tag{2}=-\int_{0}^{\sqrt{3}}\frac{t\log t}{1+t^2}\,dt$$ but: \begin{eqnarray*} \int\frac{t\log ... 4 Integrating f(z)=\dfrac{e^{iaz^2}}{z+b} along [0,R]\cup Re^{i[0,\pi/2]}\cup i[R,0] gives \begin{align} \int^\infty_0\frac{\sin(ax^2)}{x+b}\ {\rm d}x &=\int^\infty_0\frac{b\cos(ay^2)-y\sin(ay^2)}{y^2+b^2}\ {\rm d}y \end{align} To compute the first integral, we consider the function \displaystyle I(a)=\int^\infty_0\frac{e^{iay^2}}{y^2+b^2}\ {\rm d}y ... 4 "What remains" is actually the only nontrivial part of \operatorname{Li}_2\left(e^{ix}\right), given by Clausen function. The fact that one can evaluate cosine series is related to the identity\operatorname{Li}_2\left(z\right)+\operatorname{Li}_2\left(z^{-1}\right)=-\frac{\ln^2\left(-z\right)}{2}-\frac{\pi^2}{6}.$$Slightly rephrasing, we can rewrite ... 3 Here is a solution only using dilogarithm identities:$$ \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) = \zeta(2) - \log z \log(1-z). \tag{1}  \operatorname{Li}_2(1-z) + \operatorname{Li}_2(1-\tfrac{1}{z}) = -\tfrac{1}{2}\log^2 z. \tag{2}  \operatorname{Li}_2(z) + \operatorname{Li}_2(-z) = \tfrac{1}{2}\operatorname{Li}_2(z^2). \tag{3} ...

3

To complete the excellent answer of @JackD'Aurizio: \begin{align*} \Im\operatorname{Li}_2\left(1+i\sqrt3\right)=\frac{\pi \ln 2}{2}+\frac{5\sqrt3}{72}\left[\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)\right], \end{align*} where $\psi_1(z)$ denotes the trigamma function.

2

Too long for a comment: Here's a little intuitive tip: What do $~\dfrac{\sin t}t~$ and $~\dfrac{\cos t}{t^2}~$ both have in common? They are even functions. So, if you notice various series or integrals whose summand or integrand belongs to this category having a nice closed form, that should not surprise you. For instance, ...

2

Ok, i will give it a shot: Writing $\int_{0}^{\infty}e^{-t(x+2)}=\frac{1}{x+2}$ and using $\Im(e^{ix})=\sin(x)$ we may reformulate the problem as follows: $$I=\Im\left[\int_0^{\infty}dte^{-2 t}\underbrace{\int_0^{\infty}dxe^{i\pi x^2-tx}}_{J(t)}\right]$$ the inner intgral $J(t)$ is quite straightforward (and also well known because it is just the laplace ...

1

The integral can be written: $$\int^{\infty}_{0}\frac{x^n}{x^{m+n+1}}dx=\int^{\infty}_{0} x^nx^{-m-n-1} dx \\ =\int^{\infty}_{0} x^{-m-1} dx=\int^{\infty}_{0} \frac{1}{x^{m+1}} dx$$ which does not converge. You may be confusing the integral you have with the Beta function, which has the representation  ...

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