Feynman integration

Similar solution with @Will McGlaughlin 's

Let

$\begin{aligned} I(u) & = \int_0^1 \frac {x^u-1}{\ln x} dx \\ \frac {\partial I(u)}{\partial u} & = \int_0^1 \frac {\ln x \cdot x^u}{\ln x} dx \\ & = \int_0^1 x^u \ dx \\ & = \frac 1{u+1} \\ \implies I(u) & = \int \frac 1{u+1} du \\ & = \ln (u+1) + C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ I(0) & = \ln (0+1)+C = 0 & \small \color{#3D99F6} \text{Note that } I(0) = 0 \implies C = 0 \\ \implies I(u) & = \ln (u+1) \\ \implies I(2) & = \ln 3 \end{aligned}$

Therefore, $a-2 = 3-2=\boxed{1}$ .

Let the integral $\int _{ 0 }^{ 1 }{ \frac { { x }^{ b }-1 }{ \ln { x } } } dx = I(b)$

$I'(b)=\int _{ 0 }^{ 1 }{ \frac { d }{ db } (\frac { { x }^{ b }-1 }{ \ln { x } } ) } dx$

$\frac { d }{ db } ({ x }^{ b }-1)=ln(x){ x }^{ b }$

Therefore $I'(b)=\int _{ 0 }^{ 1 }{ { x }^{ b } } dx$

$I'(b)={ \frac { { x }^{ b+1 } }{ b+1 } }_{ to\quad x=0 }^{ from\quad x=1 }$

$I'(b)=\frac { 1 }{ b+1 }$

Taking the integral of both sides we reach;

$I(b)=\ln { (b+1) } +c$

We know that $\int _{ 0 }^{ 1 }{ \frac { { x }^{ b }-1 }{ \ln { x } } } dx = I(b)=\ln { (b+1) } +c$

Let $b=0$

We get $\int _{ 0 }^{ 1 }{ \frac { { x }^{ 0 }-1 }{ \ln { x } } } dx = I(0)=\ln { (0+1) } +c$

$0=\ln { (1) } +c$

$c=0$

$I(b)=\ln { (b+1) }$

$I(2)=\ln { (2+1) }$

$I(2)=\ln { (3) }$

$a=3$

$a-2=\boxed{1}$

Alternatively to the differentiation under the integral sign technique, we can convert this to a double integral. Notice $\int_0^2 x^t \,dt = \frac{x^2 - 1}{\ln(x)}.$ Let $I$ denote the integral in question, we can rewrite $I$ equivalently as $I = \int_0^1 \int_0^2 x^t \,dt \, dx$ and we can utilize Fubini's Theorem to interchange the order of integration to give $I = \int_0^2 \int_0^1 x^t\,dx\,dt$ which gives $I = \int_0^2 \frac{1}{t+1}\,dt = \ln(3)$ thus $\boxed{ a-2 = 1}.$