$$ \int_0^\infty \frac{t}{e^t} dt $$ The above integral can be computed to be exactly 1 with integration by parts. However, i just wanted to try different integration techniques to see if i could arrive at the same result.
So i carry the following substitution $ t = \ln(k) $ and end up with the following integral $$ \int_{-\infty} ^\infty \frac{\ln(k)}{k^2} dk $$ I plug this integral into WolframAlpha and i am told that it does not converge.
Please note that i would never do this substitution under normal circumstances i am just messing around with integrals.
So my question is, where did i make the mistake? Did i mess up my integration boundaries?
One way of evaluating this integral is noting that $$\Gamma(n)=\int_0^\infty t^{n-1}e^{-t}dt$$ And for non-negative integer $n$, $$\Gamma(n)=(n-1)!$$ Which gives your integral $$\Gamma(2)=1!=1$$ QED