# Definite integral challenge...

#### DreamWeaver

##### Well-known member
For $$\displaystyle m \in \mathbb{Z}^+$$, and $$\displaystyle a, \, z \in \mathbb{R} > 0$$, evaluate the definite integral:

$$\displaystyle \int_0^z\frac{x^m}{(a+\log x)}\,dx$$

[I'll be adding a few generalized forms like this in the logarithmic integrals thread, in Maths Notes, shortly... ]

#### Random Variable

##### Well-known member
MHB Math Helper
You can find an antiderivative in terms of the exponential integral.

$\displaystyle \int \frac{x^{m}}{a+\ln x} \ dx = \int \frac{e^{u(m+1)}}{a+u} \ du = e^{-a(m+1)} \int \frac{e^{m(w+1)}}{w} \ dw$

$\displaystyle = e^{-a(m+1)} \ \text{Ei} \Big( w(m+1) \Big) + C$

$\displaystyle = e^{-a(m+1)} \ \text{Ei} \Big( (a+u) (m+1) \Big) + C$

$\displaystyle = e^{-a(m+1)} \ \text{Ei} \Big( (a+\ln x) (m+1) \Big) + C$

And I think you need more restrictions on the parameters to guarantee convergence.