<s>[INST] Prove $\int_0^1\frac{\text{Li}_2(-x^2)}{\sqrt{1-x^2}}\,dx=\pi\int_0^1\frac{\ln\left(\frac{2}{1+\sqrt{1+x}}\right)}{x}\,dx$ I managed here to prove $$\int_0^1\frac{\text{Li}_2(-x^2)}{\sqrt{1-x^2}}\,dx=\pi\int_0^1\frac{\ln\left(\frac{2}{1+\sqrt{1+x}}\right)}{x}\,dx$$
but what I did was converting the LHS integral to a series then converting the series to the RHS integral. Is it possible to relate the two integrals without going through the series?
 [/INST] First consider the following definition:
$$\operatorname{Li}_2\left(x\right)=-\int _0^x\frac{\ln \left(1-t\right)}{t}\:dt.$$
This means that:
$$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:dx=-\int _0^1\frac{1}{\sqrt{1-x^2}}\underbrace{\int _0^{-x^2}\frac{\ln \left(1-t\right)}{t}\:dt}_{y=-\frac{t}{x^2}}\:dx$$
$$=-\int _0^1\frac{1}{y}\int _0^1\frac{\ln \left(1+yx^2\right)}{\sqrt{1-x^2}}\:dx\:dy.$$
Now if we use differentiation under the integral sign for that integral we obtain:
$$=-\int _0^1\frac{1}{y}\left(\pi \ln \left(\frac{1+\sqrt{1+y}}{2}\right)\right)\:dy.$$
Therefore:
$$\boxed{\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:dx=\pi \int _0^1\frac{\ln \left(\frac{2}{1+\sqrt{1+y}}\right)}{y}\:dy.}$$
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