The first value is the integral, the second is the error. You have (1.8 +- 3.3) 10^-126. That is, your integral is almost zero.

 

That is inside the bracket Scipy said. And I wouldn't even trust them to be correct without doing some checking.

Numbers so small are numerically non trustworthy. Your function is:

E^(A - B log(h) - C log^2(h)) = E^A * h^B' * h ^ (C' log(h))

where B' and C' depend on your numbers. E^A is already 10^-507 (are you sure this numbers are correct?), so if you strip it out of your integral, you have a simpler expression, and more reasonable values. This is what is killing your integral and sending it to 0.

Now, you have to normalise the values. So,  changing variables: h-> sx, dh = sdx:

sx^B' * (sx)^[C' (log(s) + log(x))]

Expand and simplify, find the value that makes more reasonable values (order of one).

Last thing: giving quad a limit in infinity can make it behave crazily. If you look at your second term, that is roughly x^(-log(x)), it decays very quickly for large values of x (10^-5 for x=20), so you can, without affecting your integral much, set a cut at a some reasonable h (plot it to be sure). You can even make an estimate comparing with the analytical result [1] of that particular piece.

Bottomline: don't use a lambda, use a function and expand your parameters. That will make the expression simpler.

def f(h):

    log10 = np.log10(h)

    B = 1936.9610182100055*5
    ...

    return ...

/David.

[1] http://www.wolframalpha.com/input/?i=Integrate[x^-log%28x%29%2C+{x%2C+k%2C+infinity}]