Use Leibnitz Theorem to solve the following-

Question

If anyone could break it down to smaller steps, I'd appreciate alot.

Answer 1

Let $$ F(a)=\int_0^{\infty}\frac{e^{-x}}{x}(1-e^{-ax})\,dx. \tag{1} $$ Differentiating $F$ with respect to $a$, we obtain $$ F'(a)=\int_0^{\infty}\frac{e^{-x}}{x}xe^{-ax}\,dx =\int_0^{\infty}e^{-(1+a)x}\,dx =\frac{1}{1+a}. \tag{2} $$ Now we integrate $(2)$ with respect to $a$: $$ F(a)=\int\frac{da}{1+a}=\ln(1+a)+C. \tag{3} $$ To determine $C$, we notice that the integral $(1)$ vanishes for $a=0$, so $$ F(0)=0=\ln(1+0)+C\implies C=0. \tag{4} $$ Therefore, we conclude that $$ \int_0^{\infty}\frac{e^{-x}}{x}(1-e^{-ax})\,dx=\ln(1+a)\qquad(a>-1). \tag{5} $$