Pie is Everywhere

The radius of the base of the bullet is $\ln(e) = 1$ and its length is $e - 1$ . So since the length of the entire hole is $e + 1$ , the volume of the hole is the volume of the bullet plus the volume of a cylinder of length $2$ and radius $1$ .

The volume of the cylindrical part of the hole is then $2\pi$ , and the volume of the bullet is

$\displaystyle\int_{1}^{e} \pi(\ln(x))^{2} dx$ .

We can solve this integral using the parts method. Let $u = (\ln(x))^{2}$ and $dv = dx$ . Then $du = 2\frac{\ln(x)}{x} dx$ and $v = x$ , and the integral then becomes, (ignoring the bounds for the moment),

$\pi x(\ln(x))^{2} - 2\pi\displaystyle\int \ln(x) dx = \pi x(\ln(x))^{2} - 2\pi(x\ln(x) - x) =$

$\pi x((\ln(x))^{2} - 2\ln(x) + 2)$ ,

where another application of the parts method was used to determine $\int \ln(x) dx$ . Evaluating this from $x = 1$ to $x = e$ gives us a volume of

$\pi e*(1 - 2 + 2) - \pi*(0 - 0 + 2) = \pi(e - 2)$ .

Adding to this the volume of the cylindrical part of the hole gives us a total volume of $e\pi$ , implying that the constant $i = \boxed{1}$ .