@DavidP There is another clean way too; ∫[0,∞] x^(c−1)·exp(−x/2) dx = 2c · ∫[0,∞] (2c·x)^(c−1)·exp(−(2c·x)/2) dx = (2c)^c · ∫[0,∞] x^(c−1)·exp(−c·x) dx = (2c)^c · ∫[0,∞] (x·exp(−x))^(c−1)·exp(−x) dx which is obviously finite because x < exp(x) for any x ≥ 0. No splitting needed!
@user21820 Hi! Thanks a lot. That seems to work nicely. The definition for gamma function is ∫[0,∞] x^(c−1)·exp(−x) dx, but then the same argument works with the factor c instead of 2c. Right?
This is the Realm of Simply Beautiful…
This is the Realm of Simply Beautiful…