Is the area independent of $n$? $ A=\int_0^\infty \frac{e^{-\frac{1}{tn}}(e^{-tn}(tn+1))}{t}~dt $ Consider the parametric equation $x=\frac{1}{t}e^{-tn}$ and $y=te^{-\frac{1}{tn}}.$ To integrate under this curve I put it in the proper form:
$$ A=\int_0^\infty \frac{e^{-\frac{1}{tn}}(e^{-tn}(tn+1))}{t}~dt $$
I noticed that by changing $n$ from 1 to 2 to 3 that the area was the same!

Is the area $A$ independent of $n?$

Do integrals like these have applications?

Letting $u=nt$ so that $du=ndt$ gives
$$A=\int_0^{\infty}\dfrac{e^{-\frac{1}{u}}(e^{-u}(u+1))}{u}du$$
which is precisely the integral when $n=1$.
