From the definition of the logarithmic integral: :$\displaystyle \operatorname {li} \left({z}\right) = \int_0^z \frac {\mathrm d t}{\ln t}$ Using Integration by Parts: {{begin-eqn}} {{eqn | l = \operatorname {li} \left({z}\right)       | r = \left[{\frac t {\ln t} }\right]_0^z - \int_0^t t \, \mathrm d (\ln^{-1} t) }} {{eqn | r = \frac z {\ln z} + \int_0^z \frac {\mathrm d t} {\ln^2 t}       | c = $\dfrac 0 {\ln 0} = 0$ and Derivative of Function to Power of Function }} {{eqn | r = \frac z {\ln z} + \frac z {\ln^2 z} + \int_0^z \frac {2 \, \mathrm d t} {\ln^3 x}       | c = Integration by Parts, $\dfrac 0 {\ln^2 0} = 0$, $t \, \mathrm d \left({ln^{-2} t}\right) = -2 \,\ln^{-3} t$ }} {{end-eqn}} This sequence can be continued indefinitely. We will consider the nature of the terms outside and inside the integral, after a number $n$ of iterations of integration by parts. Let $u_n$ be the term outside the integral. Let $v_n$ be the term inside the integral. After $n$ iterations of Integration by Parts as above, we have: :$\displaystyle \operatorname {li} \left({z}\right) = u_n + \int_0^z v_n \, \mathrm d t$ :$\displaystyle u_0 = 0$ :$\displaystyle v_0 = \frac 1 {\ln t}$ It follows that: {{begin-eqn}} {{eqn | l = \operatorname {li} \left({z}\right)       | r = u_n + \left[{t \, v_n}\right]_0^z - \int_0^z t \, \mathrm d \left({v_n}\right)       | c =  }} {{eqn | r = u_n + \left[{t \, v_n}\right]_0^z + \int_0^z -t \, \mathrm d \left({v_n}\right)       | c =  }} {{end-eqn}} <!-- tidied to here --> {{improve|This is actually a proof by induction using different language.}} which gives us the recurrence relations: {{begin-eqn}} {{eqn | n = 1       | l = u_{n + 1}       | r = u_n + \left[{t \, v_n}\right]_0^z }} {{eqn | n = 2       | l = v_{n + 1}       | r = -t \cdot \frac {\mathrm d} {\mathrm d t} \left({v_n}\right) }} {{end-eqn}} * By recurrence on $n$, with the following recurrence hypothesis: {{begin-eqn}} {{eqn | n = \text{R.H.}       | l = v_n       | r = \frac {n!} {\ln^{n + 1} t} }} {{end-eqn}} When $n = 0$, we have: :$\displaystyle v_0 = \frac 1 {\ln t} = \frac {0!} {\ln^{0 + 1} t}$ which verifies the hypothesis. By supposing true at $n$, we have at $n+1$: {{begin-eqn}} {{eqn | l = v_{n+1}       | r = -t \cdot \frac {\mathrm d} {\mathrm d t} \left({v_n}\right)       | c = from $(2)$ }} {{eqn | r = -t \cdot \frac {\mathrm d} {\mathrm d t} \left({\frac {n!} {\ln^{n + 1} t} }\right)       | c = from $(\text{R.H.})$ }} {{eqn | r = -t \cdot n! \cdot -\left({n + 1}\right) \cdot \frac 1 t \cdot \ln^{-\left({n + 1}\right) - 1} t       | c = Derivative of Function to Power of Function, Derivative of Natural Logarithm Function }} {{eqn | r = \frac {\left({n + 1}\right)!} {\ln^{n + 2} t}       | c = Definition of Factorial }} {{end-eqn}} So $(\text{R.H.})$ is verified at $n + 1$ if it is verified at $n$. So it is proved for every $n \in \N$ (since it is true at $n=0$): {{begin-eqn}} {{eqn | n = 3       | l = v_n       | r = \frac {n!} {\ln^{n+1} t} }} {{end-eqn}} By taking $(1)$, and inserting $(3)$ in, a new expression for $u_{n + 1}$ in function of $u_n$ (recursive expression): {{begin-eqn}} {{eqn | l = u_{n + 1}       | r = u_n + \left[{t \cdot \frac {n!} {\ln^{n + 1} t} }\right]_0^z       | c =  }} {{eqn | r = u_n + \frac {z \, n!} {\ln^{n + 1} z} - \frac {0 \cdot n!} {\ln^{n + 1} 0}       | c =  }} {{eqn | r = u_n + \frac {z \, n!} {\ln^{n + 1} z}       | c =  }} {{end-eqn}} That is, we can write by expanding: {{begin-eqn}} {{eqn | n = 4       | l = u_{n + 1}       | r = \sum_{i \mathop = 0}^n \frac {z \, i!} {\ln^{i + 1} z} }} {{end-eqn}} {{qed}} Category:Special Functions hmky5tx03w1lmdnc7fpyhoqq3ziixc5