Let's do some calculus! (57)

A simple use of Gamma Function along with telescopic series....... !! A nice question btw @Tapas Mazumdar

Let's write this logarithm's argument in the following manner:

$\int_{0}^{\infty} \frac{e^{-x/t}}{t} \cdot \Sigma_{i=2}^{\infty} \frac{x^{i-1}}{(i-1)(i-1)!} dx$

which the infinite series can be expressed as:

$\Sigma_{i=2}^{\infty} \frac{x^{i-1}}{(i-1)(i-1)!} = \Sigma_{i=2}^{\infty} [\int_{0}^{x} \frac{y^{i-2}}{(i-1)!} dy] = \int_{0}^{x} \frac{1}{y} \cdot \Sigma_{i=2}^{\infty} \frac{y^{i-1}}{(i-1)!} dy = \int_{0}^{x} \frac{e^{y} - 1}{y} dy$ .

Returning to our original integral, we now have the double integral in $x$ and $y$ :

$\int_{0}^{\infty} \int_{0}^{x} \frac{e^{-x/t}}{t} \cdot \frac{e^{y}-1}{y} dy dx$

and by reversing the order of integration, we can now compute this as:

$\int_{0}^{\infty} \int_{y}^{\infty} \frac{e^{-x/t}}{t} \cdot \frac{e^{y}-1}{y} dx dy$ ;

or $\int_{0}^{\infty} \frac{e^{y}-1}{y} \cdot e^{-y/t} dy;$

or $\int_{0}^{\infty} \frac{e^{-(1/t - 1)y} - e^{-(1/t)y}}{y} dy;$

or $ln(\frac{1/t}{1/t -1}) = ln(\frac{1}{1-t}) = -ln(1-t).$

Ultimately, we have the expression $ln[-ln(1-t)]$ in which the argument must be positive in order to be valid. Hence:

$-ln(1-t) > 0 \Rightarrow ln(1-t) < 0$ ;

or $1-t > 0 \Rightarrow 1 > t$ AND $1-t < 1 \Rightarrow 0 < t$

which leaves $\boxed{t \in (0,1)}$ as the required interval for $t.$