I am currently trying to find the primitive of the following function:
$$f(t)= \int \frac{a}{\ln(at+b)}dt$$
Where $a$ and $b$ are constants. Furthermore $a>0$ and $0<b<1.$ Additionally, $a$ is of inverse time units and $b$ is dimensionless.
I tried using differentiation under the integral sign by introducing:
$$I(t)= \int \frac{a(at+b)^x}{\ln(at+b)}dt$$
Differentiating it with respect to $x$ and then find $I’$ and substitute by $x=0$ however i get stuck because the result that i get isn’t dimensionless as it should be (probably because of an error of mine somewhere). Could someone help please? Thank you.
Making the following change of variable $x=\ln(at+b)$ you get $e^x dx=a dt$ and your primitive transforms into: $$\int \frac{e^x}{x} dx$$ The change of variable maps from $t\in[0,\infty)$ into $x\in [\log b,\infty)$. Define your primitive as the one vanishing at $x=\log b$ by $$F(x)=\int_{\log b}^x \frac{e^u}{u} du$$
Then you get for any $x\in \mathbb R_*$,
$$F(x)= \int_{-\infty}^x \frac{e^u}{u} du - \int_{-\infty}^{\log b} \frac{e^u}{u} du$$ which is by definition, $$F(x)= \operatorname{Ei}(x)- \operatorname{ Ei}(\log b)$$
where $\operatorname{Ei}(x)$ is known as the exponential integral defined on $\mathbb R^*$. Going back to your initial coordinates you get the primitive vanishing at $t=0$: $$f(t)= F(\ln(at+b))=\operatorname{Ei}(\ln(at+b))-\operatorname{Ei}(\log b)$$
defined for $t\in[0,\infty)$ except at point $t=\frac{1-b}{a}$. You can add any constant to this primitive if you wish.