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Prove some properties of the integral logarithm, Li (x)

by RoRi

The integral logarithm \operatorname{Li}(x) is defined for x \geq 2 by

    \[ \operatorname{Li}(x) = \int_2^x \frac{dt}{\log t}. \]

Prove the following properties of \operatorname{Li}(x).

  1. \displaystyle{ \operatorname{Li}(x) = \frac{x}{\log x} + \int_2^x \left(\frac{dt}{\log^2 t}\right) - \frac{2}{\log 2}}.
  2. \displaystyle{ \operatorname{Li}(x) = \frac{x}{\log x} + \sum_{k=1}^{n-1} \left(\frac{k!x}{\log^{k+1} x}\right) + n! \int_2^x \left( \frac{dt}{\log^{n+1} t} \right) + C_n} where C_n is a constant depending on n. Find the value of C_n for each n.
  3. Prove there exists a constant b such that

        \[ \operatorname{Li}(x) = \int_b^{\log x} \frac{e^t}{t} \, dt \]

    and find the value of this constant.

  4. Let c = 1 + \frac{1}{2} \log 2. Find an expression for

        \[ \int_c^x \frac{e^{2t}}{t-1} \, dt \]

    in terms of \operatorname{Li}(x).

  5. Define a function for x > 3 by

        \[ f(x) = e^4 \operatorname{Li}(e^{2x-4}) - e^2 \operatorname{Li}(e^{2x-2}). \]

    Prove that

        \[ f'(x) = \frac{e^{2x}}{x^2-3x+2}. \]

  1. Proof. We derive this by integrating by parts. Let

        \begin{align*} u &= \frac{1}{\log t} & du &= \frac{-1}{t \log^2 t} \, dt \\ dv &= dt & v &= t. \end{align*}

    Then we have

        \begin{align*} \operatorname{Li}(x) &= \int_2^x \frac{dt}{\log t} \\[9pt] &= \frac{t}{\log t} \Bigr \rvert_2^x + \int_2^x \frac{1}{\log^2 t} \, dt \\[9pt] &= \frac{x}{\log x} + \int_2^x \frac{dt}{\log^2 t} - \frac{2}{\log 2}. \qquad \blacksquare \end{align*}

  2. Proof. The proof is by induction. Starting with part (a) we have

        \[ \operatorname{Li}(x) = \frac{x}{\log x} + \int_2^x \frac{dt}{\log^2 t} - \frac{2}{\log 2}. \]

    To evaluate the integral in this expression we integrate by parts with

        \begin{align*} u &= \frac{1}{\log^2 t} & du &= \frac{-2}{t \log^3 t} \, dt \\ dv &= dt & v &= t. \end{align*}

    This gives us

        \begin{align*} \int_2^x \frac{dt}{\log^2 t} &= \frac{t}{\log^2 t} \Bigr \rvert_2^x + 2 \int_2^x \frac{dt}{\log^3 t} \\[9pt] &= \frac{x}{\log^2 x} + 2 \int_2^x \frac{dt}{\log^3 t} - \frac{2}{\log^2 2}. \end{align*}

    Therefore we have

        \begin{align*} \operatorname{Li}(x) &= \frac{x}{\log x} + \int_2^x \frac{dt}{\log^2 t} - \frac{2}{\log 2} \\[9pt] &= \frac{x}{\log x} + \frac{x}{\log^2 x} + 2 \int_2^x \frac{dt}{\log^3 t} - \left( \frac{2}{\log 2} + \frac{2}{\log^2 2} \right) \\[9pt] &= \frac{x}{\log x} + \sum_{k=1}^{n-1} \frac{n! x}{\log^{k+1} x} + n! \int_2^x \frac{dt}{\log^{n+1} t} + C_n. \end{align*}

    where C_n = -2 \sum_{k=1}^n \frac{(k-1)!}{\log^k 2}. This is the case n = 2. Now, assume the formula hold for some integer m \geq 2. Then we have

        \[ \operatorname{Li}(x) &= \frac{x}{\log x} + \sum_{k=1}^{m-1} \frac{k! x}{\log^{k+1} x} + m! \int_2^x \frac{dt}{\log^{m+1} t} - 2 \sum_{k=1}^m \frac{(k-1)!}{\log^k 2}.\]

    We then evaluate the integral in this expression using integration by parts, as before, let

        \begin{align*} u &= \frac{1}{\log^{m+1} t} & du &= \frac{-(m+1)}{t \log^{m+2} t} \\ dv &= dt & v &= t. \end{align*}

    Therefore, we have

        \begin{align*} \int_2^x \frac{dt}{\log^{m+1} t} &= \frac{t}{\log^{m+1} t} \Bigr \rvert_2^x + (m+1) \int_2^x \frac{dt}{\log^{m+2} t} \\[9pt] &= \frac{x}{\log^{m+1} x} + (m+1) \int_2^x \frac{dt}{\log^{m+2} t} - \frac{2}{\log^{m+1} 2}. \end{align*}

    Plugging this back into the expression we had from the induction hypothesis we obtain

        \begin{align*} \operatorname{Li}(x) &= \frac{x}{\log x} + \sum_{k=1}^{m-1} \frac{k! x}{\log^{k+1} x} + m! \int_2^x \frac{dt}{\log^{m+1} t} - 2 \sum_{k=1}^m \frac{(k-1)!}{\log^k 2} \\[10pt] &= \frac{x}{\log x} + \sum_{k=1}^{m-1} \frac{k! x}{\log^{k+1} x} + m! \left( \frac{x}{\log^{m+1} x} + (m+1) \int_2^x \frac{dt}{\log^{m+2} t} - \frac{2}{\log^{m+1} 2} \right) \\ & \qquad - 2 \sum_{k=1}^m \frac{(k-1)!}{\log^k 2} \\[10pt] &= \frac{x}{\log x} + \left( \sum_{k=1}^{m-1} \frac{k! x}{\log^{k+1} x} + \frac{m! x}{\log^{m+1}} \right) + (m+1)! \int_2^x \frac{dt}{\log^{m+2} t} \\ & \qquad - 2 \left( \frac{m!}{\log^{m+1} 2} + \sum_{k=1}^m \frac{(k-1)!}{\log^k 2} \right) \\[10pt] &= \frac{x}{\log x} + \sum_{k=1}^m \frac{k! x}{\log^{k+1} x} + (m+1)! \int_2^x \frac{dt}{\log^{m+2} t} - 2 \sum_{k=1}^{m+1} \frac{(k-1)!}{\log^k 2} \\[10pt] &= \frac{x}{\log x} + \sum_{k=1}^m \frac{k! x}{\log^{k+1} x} + (m+1)! \int_2^x \frac{dt}{\log^{m+2} t} + C_{m+1}. \end{align*}

    Therefore, the formula holds for the case m+1, and hence, for all integers n \geq 2, where

        \[ C_n = -2 \sum_{k=1}^n \frac{(k-1)!}{\log^k x}. \qquad \blacksquare \]

  3. Proof. We start with the definition of the integral logarithm,

        \[ \operatorname{Li}(x) = \int_2^x \frac{dt}{\log t} \]

    and make the substitution s = \log t, ds = \frac{dt}{t}. This gives us dt = t \, ds = e^s \, ds. Therefore,

        \begin{align*} \operatorname{Li}(x) &= \int_2^x \frac{dt}{\log t} \\[9pt] &= \int_{\log 2}^{\log x} \frac{e^s}{s} \, ds \\[9pt] &= \int_b^{\log x} \frac{e^t}{t} \, dt \end{align*}

    where b = \log 2 is a constant. \qquad \blacksquare

  4. (Note: In the comments, tom correctly suggests an easier way to do this is to use part (c) along with translation and expansion/contraction of the integral. The way I have here works also, but requires an inspired choice of substitution.) We start with the given integral,

        \[ \int_c^x \frac{e^{2t}}{t-1} \, dt \]

    and make the substitution

        \begin{align*} s &= e^{2t-2} & \implies && t &= 1 + \frac{1}{2} \log s \\ ds &= 2e^{2t-2}\, dt & \implies && dt &= \frac{1}{2} e^{2-2t} \, ds = \frac{ds}{2s}. \end{align*}

    Therefore, using the given fact that c = 1 + \frac{1}{2} \log 2, we have

        \begin{align*} \int_c^x \frac{e^{2t}}{t-1} \, dt &= \int \limits_{e^{2(1+\frac{1}{2}\log 2) - 2}}^{e^{2x-2}} \frac{e^2 s}{\frac{1}{2} \log s} \frac{ds}{2s} \\[10pt] &= e^2 \int_2^{e^{2x-2}} \frac{ds}{\log s} \\[10pt] &= e^2 \operatorname{Li}(e^{2x-2}). \end{align*}

  5. From part (d) we know that

        \[ e^2 \operatorname{Li}(e^{2x-2}) = \int_c^x \frac{e^{2t}}{(t-1)} \, dt. \]

    Then, for the term e^4 \operatorname{Li}(e^{2x-4}) we consider the integral

        \[ \int_d^x \frac{e^{2t}}{t-2} \, dt, \]

    where d = 2 + \frac{1}{2} \log 2. Similar to part (d) we make the substitution,

        \begin{align*} s &= e^{2t-4} & \implies && t &= 2 + \frac{1}{2} \log s \\ ds &= 2e^{2t-4} \, dt & \implies && dt &= \frac{1}{2}e^{4-2t} \, ds = \frac{ds}{2s}. \end{align*}

    This gives us

        \begin{align*} \int_d^x \frac{e^{2t}}{t-2} &= \int \limits_{e^{2(2+\frac{1}{2} \log 2) - 2}}^{e^{2x-4}} \frac{e^4 s}{\frac{1}{2} \log s} \frac{ds}{2s} \\[9pt] &= e^4 \int_2^{e^{2x-4}} \frac{ds}{\log s} \\[9pt] &= e^4 \operatorname{Li}(e^{2x-4}). \end{align*}

    Therefore, we have

        \[ f(x) = e^4 \operatorname{Li}(e^{2x-4}) - e^2 \operatorname{Li}(e^{2x-2}) = \int_d^x \frac{e^{2t}}{t-2} \,dt - \int_c^x \frac{e^{2t}}{t-1} \, dt. \]

    Taking the derivative we then have

        \begin{align*} f'(t) &= \frac{e^{2x}}{x-2} - \frac{e^{2x}}{x-1} \\ &= \frac{e^{2x}(x-1) - e^{2x}(x-2)}{(x-2)(x-1)} \\ &= \frac{e^{2x}}{x^2 - 3x + 2}. \qquad \blacksquare \end{align*}

4 comments

  1. Artem says:

    Part (d) can be done naturally with 2 substitutions: first one u = t – 1. After this substitution the integral will resemble the integral in part (c) – from where we should get an inspiration to do the second substitution to make the integral look like in (c), and the conclude the result

    Part (e) can be done with the 1st theorem of calculus. 1/log(t) is continuous for the positive t, so the derivative of Li(x) at e^{2x – 2} is simply 1/log(e^{2x-2}). Then the chain rule is used.

  2. tom says:

    Part (d) can be done using the translation and expansion properties along with part (c). Your substitution was interesting, but at this stage might have eluded me.

      • RoRi says:

        Oh, there was also a typo in part (d). I had something an integral of \frac{ds}{s} instead of \frac{ds}{\log s}. Fixed now. Anyway, you’re right doing this using part (c) is easier. I’m not sure why I went with this substitution… it is not an obvious one to make at all. (I’m sure that I chose it because of the value of the constant c we are given in the problem.) I’ll put in a note directing people to your comment, but I’ll leave the substitution solution up there since it’s still correct and maybe people will like to see an overly complicated way to do it.

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