Laplace transform using numerical integration in Python has very poor precision

I have written a function to compute the Laplace transform of a function using scipy.integrate.quad. It is not a very sophisticated function and currently performs poorly on the probability density function of an Erlang distribution.

I have included all my work below. I first compute the Laplace transform and then the inverse in order to compare it to the original p.d.f. of the Erlang. I use mpmath for this. The mpmath.invertlaplace is not the problem as it manages to convert the closed-form Laplace transform back to the original p.d.f. quite perfectly.

Please help me understand what the problem is with my numerical laplace transform. I get the following error but have not been able to resolve it.

IntegrationWarning: The occurrence of roundoff error is detected, which prevents the requested tolerance from being achieved. The error may be underestimated. a=0,b=np.inf,limit=limit)[0]

PLOT

CODE

import numpy as np
import matplotlib.pyplot as plt
import math as m
import mpmath as mp
import scipy.stats as st
from scipy.integrate import quad


def get_laplace(func,limit=10000):
    
    '''
    Returns laplace transfrom function
    '''
    
    def laplace(s):
        '''Numerical laplace transform'''
        # Seperate into real and imaginary parts
        x = np.real(s)
        y = np.imag(s)
        
        def real_func(t):
            return m.exp(-x*t)*m.cos(y*t)*func(t)
        
        def imag_func(t):
            return m.exp(-x*t)*m.sin(y*t)*func(t)
        
        real_integral = quad(real_func,
                             a=0,b=np.inf,limit=limit)[0]
        
        imag_intergal = quad(real_func,
                             a=0,b=np.inf,limit=limit)[0]
        
        return complex(real_integral,-imag_intergal)
    
    return laplace
        

def L_erlang(s,lam,k):
    '''
    Closed form laplace transform of Erlang or Gamma distribution.
    '''
    return (lam/(lam+s))**k


if __name__ == '__main__':
    
    # Setup Erlang probability density function
    k = 5
    lam = 1
    pdf = st.erlang(a=k,scale=1/lam).pdf
    
    # Laplace transforms
    Lnum = get_laplace(pdf)          # numerical approximation
    L = lambda s: L_erlang(s,lam,k)  # closed form
    
    
    # Use mpmath library to perform inverse laplace
    
    # Invserse transfrom on numerical laplace function
    invLnum = lambda t: mp.invertlaplace(Lnum,t,
                                         method='dehoog',
                                         dps=5, 
                                         degree=3)
    
    # Invserse transfrom on closed-form laplace function
    invL = lambda t: mp.invertlaplace(L,t,
                                      method='dehoog',
                                      dps=5, 
                                      degree=3)
    
    # Grid to visualise
    T = np.linspace(0.1,10,10)
    # Get values of inverse transforms
    lnum = np.array([invLnum(t) for t in T])
    l = np.array([invL(t) for t in T])
    
    # Plot
    plt.plot(T,lnum,label='from numerical laplace')
    plt.plot(T,l,label='from closed-form laplace')
    plt.plot(T,pdf(T),label='original pdf',linestyle='--')
    plt.legend(loc='best')
    plt.show()

UPDATE

After two cups of VERY strong coffee, I managed to see the obvious mistake and make the code work. It quite embarrassing actually. Have a look at this line of code:

imag_intergal = quad(real_func,
                     a=0,b=np.inf,limit=limit)[0]

Hmmm, real_func hey? So it should read:

imag_intergal = quad(imag_func_func,
                     a=0,b=np.inf,limit=limit)[0]

The one gets this lovely plot:

Conclusion So why go through all the trouble to numerically perform the Laplace transform of something that we have a closed form solution for. That is because the interest lies somewhere else. We do not have a closed expression for the conditional future lifetime distribution which is similar to the hazard function. Lets call it h. Then for the Erlang distribution erl = st.erlang(a=k,scale=1/lam) that has been active for tau units of time we have h = lambda t: erl.pdf(t+tau)/erl.sf(tau). This distribution can be used as a holding time in a Semi-Markov Model (SMP). To analyse the transient behaviour of the SMP one use Laplace transforms. Usually only pdfs are used but now I can use hazard functions. Its pretty cool because it means one can model transient behaviour without assuming everything is new.