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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*}