A calculus problem by rahul singh

$\begin{aligned} I(a) & = \int_0^1 \frac {x^a-1}{\ln x} dx & \small \color{#3D99F6} \text{Let }x = e^u \implies dx = e^u du \\ & = \int_{-\infty}^0 \frac {e^u(e^{au}-1)}u du \\ \frac {\partial I(a)}{\partial a} & = \int_{-\infty}^0 e^{(a+1)u} du = \frac {e^{(a+1)u}}{a+1} \bigg|_{-\infty}^0 = \frac 1{a+1} \\ \implies I(a) & = \int \frac 1{a+1} da = \ln (a+1) + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ I(0) & = \ln 1 + C = 0 & \small \color{#3D99F6} \implies C = 0 \\ \implies I(a) & = \ln (a+1) \\ I(7) & = \ln 8 \approx \boxed{2.079} \end{aligned}$

$\int \frac{x^7-1}{\log (x)} \, dx \Rightarrow \text{Ei}(8 \log (x))-\text{li}(x)$

$\text{Ei}(8 \log (0))-\text{li}(0) \Rightarrow 0$

$\underset{x\to 1}{\text{lim}}(\text{Ei}(8 \log (x))-\text{li}(x)) \Rightarrow \log (8) \approx 2.07944154167984$