Find the Laplace Transform of given function 1√t?

Let f(t) be a function of ‘t’ defined for 0 ≤ t≤ ∞, then the Laplace Transform of f(t) denoted by L f(t) or F(s), is defined by
L f(t) = F(s)= ^­­∫₀∞e^-stf(t)dt, provided the integral exist for all s larger than or equal to some value ‘s₀’. The parameter s appearing in (1) is a real or Complex Number. In general ‘s’ is taken to be positive real number. The symbol ‘L’ appearing in (1) is known as Laplace transform operator and the function e^-st appearing in the integral is called the kernel transform.
Let f(t)= 1√t then f(t)→∞ as t→0. Thus f(t) is not piecewise continuous in every finite interval in The Range t≥0, but still its Laplace transform exists. By the definition of Laplace transform, we have
L f(t) = F(s)= ^­­∫₀∞e^-stf(t)dt,
Therefore for f(t)= 1√t,
L1√t=∫₀∞e^-st1√tdt,
L1√t=∫₀∞e^-x x^-1/2 dx,[on putting st=x=dt=dx/s],
L1√t=1/√s∫₀∞e^-x x^(1/2)-1dx,
L1√t=1/√s_­­­­­­­­­­­­_­ gamma function (1/2), [by definition of gamma function ∫e^-uu^n-1du=gamma (n)]
L1√t=√π/s.
Thus the condition stated in Theorem 1 is not necessary for the existence of the Laplace Transform.