Tag: Integrals

Prove an integral formula for ∑ (sin (nx)) / n²

by RoRi

March 31, 2016

Prove that the series

$\sum_{n=1}^{\infty} \frac{\sin (nx)}{n^2}$

converges for all $x \in \mathbb{R}$ and let $f(x)$ denote the value of this sum for each $x$ . Prove that $f(x)$ is continuous for $x \in [0, \pi]$ and prove that

$\int_0^{\pi} f(x) \, dx = 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^3}.$

Proof. First, the series converges for all real $x$ by the comparison test since

$|\sin (nx)| \leq 1 \quad \implies \quad \left|\frac{\sin (nx)}{n^2}\right| \leq \frac{1}{n^2}$

for all $n$ . Therefore, the convergence of $\sum \frac{1}{n^2}$ implies the convergence of $\sum \frac{\sin (nx)}{n^2}$ . Furthermore, this convergence is uniform by the Weierstrass $M$ -test with $M_n$ given by $\frac{1}{n^2}$ , and again $\sum \frac{1}{n^2}$ converges. Thus, by Theorem 11.2 (page 425 of Apostol),

$f(x) = \sum_{n=1}^{\infty} \frac{\sin (nx)}{n^2}$

is continuous on the interval $[0, \pi]$ . Therefore, we may apply Theorem 11.4 (page 426 of Apostol):

$\begin{align*} && \sum_{k=1}^{\infty} \int_0^{\pi} \frac{\sin (kt)}{k^2} \, dt &= \int_0^{\pi} \sum_{k=1}^{\infty} \frac{ \sin (kt)}{t^2} \, dt \\[9pt] \implies && \sum_{k=1}^{\infty} \left( \frac{-\cos (kt)}{k^3} \Bigr \rvert_0^{\pi} \right) &= \int_0^{\pi} f(x) \, dx \\[9pt] \implies && \int_0^{\pi} f(x) \, dx &= 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^3} \end{align*}$

since $\cos 0 - \cos (k \pi) = 0$ if $k = 2n$ and equals 2 if $k = 2n-1. \qquad \blacksquare$

Show that we cannot interchange a limit and integral for f_n(x) = nxe^-nx²

by RoRi

March 31, 2016

For each positive integer $n$ define

$f_n (x) = nxe^{-nx^2} \qquad x \in \mathbb{R}.$

Prove that the following limit and integral cannot be interchange:

$\lim_{n \to \infty} \int_0^1 f_n (x) \, dx \neq \int_0^1 \left( \lim_{n \to \infty} f_n (x) \, dx \right).$

Proof. First, we have

$\begin{align*} \lim_{n \to \infty} \left( \int_0^1 f_n (x) \, dx \right) &= \lim_{n \to \infty} \left( \int_0^1 nxe^{-nx^2} \, dx \right) \\[9pt] &= \lim_{n \to \infty} \left( -\frac{1}{2} e^{-nx^2} \Bigr \rvert_0^1 \right) \\[9pt] &= \lim_{n \to \infty} \left( -\frac{1}{2} \left( e^{-n} - 1 \right) \right) \\[9pt] &= \frac{1}{2}. \end{align*}$

On the other hand,

$\begin{align*} \int_0^1 \left( \lim_{n \to \infty} f_n (x) \right) \, dx &= \int_0^1 \left( \lim_{n \to \infty} nxe^{-nx^2} \right) \, dx \\[9pt] &= \int_0^1 0 \, dx \\[9pt] &= 0. \end{align*}$

Hence,

$\lim_{n \to \infty} \int_0^1 f_n (x) \, dx \neq \int_0^1 \lim_{n \to \infty} f_n (x) \, dx. \qquad \blacksquare$

Prove an inequality relating sums and integrals

by RoRi

March 4, 2016

Prove that

$\sum_{k=1}^{n-1} f(k) \leq \int_1^n f(x) \, dx \leq \sum_{k=2}^n f(k)$

where $f$ is a nonnegative, increasing function defined for all $x \geq 1$ .

Use this to deduce the inequalities

$e n^n e^{-n} < n! < e n^{n+1} e^{-n}$

by taking $f(x) = \log x$ .

Proof. Since $f(x)$ is increasing we know $f([x]) \leq f(x) \leq f([x+1])$ (where $[x]$ denotes the greatest integer less than or equal to $x$ ) for all $x \geq 1$ . Define step functions $s$ and $g$ by

$s(x) = f([x]), \qquad t(x) = f([x+1]).$

Then $s$ and $t$ are constant on the open subintervals of the partition

$P = \{ 0,1,2, \ldots, n \}.$

So,

$\begin{align*} \int_1^n s(x) \, dx = \sum_{k=1}^{n-1} s_k (x_k - x_{k-1}) = \sum_{k=1}^{n-1} f(k) \\[9pt] \int_1^n t(x) \, dx = \sum_{k=1}^{n-1} t_k (x_k - x_{k-1}) = \sum_{k=1}^{n-1} f(k+1) = \sum_{k=2}^n f(k). \end{align*}$

Since $f$ is integrable we must have

$\int_1^n s(x) \, dx \leq \int_1^n f(x) \, dx \leq \int_1^n t(x) \, dx$

for every pair of step functions $s \leq f \leq t$ . Hence,

$\sum_{k=1}^{n-1} f(k) \leq \int_1^n f(x) \, dx \leq \sum_{k=2}^n f(k). \qquad \blacksquare$

Next, if we take $f(x) = \log x$ (which is nonnegative and increasing on $x \geq 1$ ) we have

$\begin{align*} &&\sum_{k=1}^{n-1} \log k \leq \int_1^n \log x \, dx \leq \sum_{k=2}^n \log k \\[9pt] \implies && \log(n-1)! \leq n \log n - n + 1 \leq \log n! \\[9pt] \implies && (n-1)! \leq e n^n e^{-n} \leq n!. \end{align*}$

From this we then get two inequalities

$n! \leq en^{n+1} e^{-n} \quad \text{and} \quad e n^n e^{-n} \leq n!.$

Therefore,

$en^n e^{-n} \leq n! \leq e n^{n+1} e^{-n}.$

Find constants so the limit of an integral has a prescribed value

by RoRi

December 25, 2015

Find values for the constants $a$ and $b$ so that

$\lim_{x \to 0} \frac{1}{bx - \sin x} \int_0^x \frac{t^2 \, dt}{\sqrt{a+t}} = 1.$

First, since the integral $\int_0^x \frac{t^2 \, dt}{\sqrt{a+t}}$ is continuous (Theorem 3.4 of Apostol) we know

$\lim_{x \to 0} \int_0^x \frac{t^2 \, dt}{\sqrt{a+t}} = \int_0^0 \frac{t^2 \, dt}{\sqrt{a+t}} = 0.$

We also have

$\lim_{x \to 0} (bx - \sin x) = 0.$

Hence, we may apply L’Hopital’s rule to the quotient giving us (using the First Fundamental Theorem of Calculus for the numerator)

$1 = \lim_{x \to 0} \frac{\int_0^x \frac{t^2 \, d}{\sqrt{a+t}}}{bx - \sin x} = \lim_{x \to 0} \frac{ \frac{x^2}{\sqrt{a+x}}}{b - \cos x}.$

Since the limit as $x \to 0$ of the numerator is 0, we must have the limit of the denominator equal to 0 as well (otherwise the whole limit would be 0 instead of 1).

$\lim_{x \to 0} (b - \cos x) = 0 \quad \implies \quad b - \cos 0 = 0 \quad \implies \quad b = 1.$

Now, substituting $b = 1$ back into our limit and using L’Hopital’s two more times (since again we have the indeterminate form $0/0$ ),

$\begin{align*} \lim_{x \to 0} \frac{\frac{x^2}{\sqrt{a+x}}}{1 - \cos x} &= \lim_{x \to 0} \frac{2x}{(\sin x)\sqrt{a+x} + \frac{1 - \cos x}{2 \sqrt{a+x}}} \\[9pt] &= \lim_{x \to 0} \frac{2}{\cos x \sqrt{a+x} + \frac{\sin x}{2 \sqrt{x+a}} + \frac{\sin x}{2 \sqrt{x+a}} - \frac{1 - \cos x}{4 \sqrt{x+a}}} \\[9pt] &= \frac{2}{\sqrt{a}}. \end{align*}$

Therefore,

$\frac{2}{\sqrt{a}} = 1 \quad \implies \quad a = 4.$

Hence, the requested constants are

$a = 4, \qquad b = 1.$

Prove or disprove given statements for functions such that f(x) = o(g(x))

by RoRi

December 22, 2015

Let $f$ and $g$ be functions, both differentiable in a neighborhood of 0, with $g(x) > 0$ and such that

$f(x) = o(g(x)) \qquad \text{as} \qquad x \to 0.$

Prove or disprove each the following statements.

$\displaystyle{ \int_0^x f(t) \, dt = o \left( \int_0^x g(t) \, dt \right)}$ as $x \to 0$ .
$f'(x) = o(g'(x))$ as $x \to 0$ .

True.
Proof. Since $f(x) = o(g(x))$ as $x \to 0$ we know by the definition of $o(g(x))$ that

$\lim_{x \to 0} \frac{f(x)}{g(x)} = 0.$

Thus, for every $\varepsilon > 0$ there exists a $\delta > 0$ such that

$|x| < \delta \quad \implies \quad \left| \frac{f(x)}{g(x)} \right| < \varepsilon.$

So, for $|x| < \delta$ we have

$\begin{align*} \left| \frac{\int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} \right| &\leq \frac{\int_0^x |f(t)| \, dt}{\left| \int_0^x g(t) \, dt \right|} \\[9pt] &< \frac{\varepsilon \int_0^x g(t) \, dt }{\left| \int_0^x g(t) \, dt \right|} \\[9pt] &= \varepsilon. \end{align*}$

The final line follows since $g > 0$ by hypothesis. Therefore,

$|x| < \delta \quad \implies \quad \left| \frac{\int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} \right| < \varepsilon.$

Hence,

$\lim_{x \to 0} \frac{ \int_0^x f(t) \, dt}{\int_0^x g(t) \, dt} = 0.$

By definition, we then have

$\int_0^x f(t) \, dt = o \left( \int_0^x g(t) \, dt \right). \qquad \blacksquare$
False.
Consider $f(x) = x^2 \sin \left( \frac{1}{x} \right)$ for $x \neq 0$ and $f(x) = 0$ for $x = 0$ . Then, for $x \neq 0$ ,

$f'(x) = 2x \sin \left( \frac{1}{x} \right) - \cos \left( \frac{1}{x} \right).$

For $x = 0$ we have $f'(0) = 0$ .

Next,

$\begin{align*} f(x) &= x^2 \sin \left( \frac{1}{x} \right) \\[9pt] &= x^2 \left( \frac{\sin \left( \frac{1}{x} \right)}{\frac{1}{x}} \right) \left( \frac{1}{x} \right) \\[9pt] &= x \left( \frac{\sin \left( \frac{1}{x} \right)}{\frac{1}{x}} \right). \end{align*}$

Since $\lim_{x \to 0} \frac{\sin (1/x)}{1/x} = 1$ we have $f(x) = o(x)$ as $x \to 0$ . However, $f'(x) \neq o(1)$ since

$\lim_{x \to 0} \left( 2 \sin \left( \frac{1}{x}\right) - \cos \left( \frac{1}{x} \right) \right)$

does not exist.

Use Taylor polynomials to approximate the integral of sin x / x

by RoRi

December 21, 2015

Using the Taylor polynomial approximation of $\sin x$ find an approximation for the integral

$\int_0^1 \frac{\sin x}{x} \, dx.$

Give an estimate for the error of the approximation. [Define $\frac{\sin x}{x} = 1$ when $x = 0$ .]

We know (from this exercise, Section 7.8, Exercise #1) that the Taylor polynomial approximation of $\sin x$ is given by

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + E_{2n}(x), \qquad \text{where} \quad |E_{2n}(x)| \leq \frac{|x|^{2n+1}}{(2n+1)!}.$

This implies

$\begin{align*} &&\frac{\sin x}{x} &= 1 - \frac{x^2}{3!} + \frac{x^4}{5!} + \frac{E_{2n}(x)}{x} \\[9pt] \implies && \int_0^1 \frac{\sin x}{x} \, dx &= \int_0^1 \left( 1 - \frac{x^2}{3!} + \frac{x^4}{5!} + \frac{E_{2n}(x)}{x} \right) \, dx \\[9pt] &&&= \left( x - \frac{x^3}{3\cdot 3!} + \frac{x^5}{5 \cdot 5!} \right) \Bigr \rvert_0^1 + \int_0^1 \frac{E_{2n}(x)}{x} \, dx \\[9pt] &&&= 0.9461 + R \end{align*}$

where

$|R| \leq \int_0^1 \frac{x^6}{7!} \, dx = \frac{1}{7\cdot 7!} \approx 0.00002.$

Prove an inequality for the integral of 1 / (1 + x⁴)

by RoRi

December 21, 2015

Prove that

$0.493948 < \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx < 0.493958.$

(Note: I cannot get the bounds Apostol asks for. I prove a different set below. I cannot figure out if it is a mistake in the book or not.)

Proof. Using the algebraic identity, valid for $0 < x < 1$ ,

$\frac{1}{1-x} = 1+x+x^2+\cdots + x^n + \frac{x^{n+1}}{1-x},$

we obtain

$\frac{1}{1+x^4} = 1 - x^4 + x^8 - x^{12} + \cdots + (-1)^n x^{4n} + (-1)^{n+1} \frac{x^{4n+4}}{1+x^4}.$

Therefore, integrating term by term,

$\begin{align*} \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx &= \int_0^{\frac{1}{2}} \left( 1 - x^4 + x^8 + \cdots + (-1)^n x^{4n} + (-1)^{n+1} \frac{x^{4n+4}}{1+x^4} \right) \, dx \\[9pt] &= \left( x - \frac{x^5}{5} + \frac{x^9}{9} - \cdots + (-1)^n \frac{x^{4n+1}}{4n+1} \right)\Bigr \rvert_0^{\frac{1}{2}} + (-1)^{n+1} \int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx. \end{align*}$

Furthermore, we have

$\int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx < \int_0^{\frac{1}{2}} x^{4n+4} \, dx = \left( \frac{1}{2} \right)^{4n+5} \left( \frac{1}{4n+5} \right).$

Taking $n = 2$ , we then have

$\int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx = \left( \frac{1}{2} \right) - \frac{(1/2)^5}{5} + \frac{(1/2)^9}{9} - \int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx.$

From the inequality for this integral we then have

$\begin{align*} &\frac{1}{2} - \frac{(1/2)^5}{5} + \frac{(1/2)^9}{9} - \left( \frac{1}{2} \right)^{4n+5} \left( \frac{1}{4n+5} \right) < \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx\\ \intertext{and} &\int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx < \frac{1}{2} - \frac{(1/2)^5}{5} + \frac{(1/2)^9}{9} \\[9pt] \implies & 0.493958 < \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx < 0.49367. \qquad \blacksquare \end{align*}$

Prove an expression for the integral from 0 to 1 of (1 + x³⁰) / (1+x⁶⁰)

by RoRi

December 21, 2015

Prove that there exists a number $c$ with $0 < c < 1$ such that

$\int_0^1 \frac{1+x^{30}}{1+x^{60}} \, dx = 1 + \frac{c}{31}.$

Proof. For $0 < x < 1$ we have

$1 < \frac{1+x^{30}}{1+x^{60}} < 1 + x^{30}.$

(Since for $0 < x < 1$ we know $1+x^{60} < 1+x^{30}$ .) Therefore, integrating the terms in the inequality from 0 to 1,

$\begin{align*} &\int_0^1 \, dx < \int_0^1 \frac{1+x^{30}}{1+x^{60}} \, dx < \int_0^1 (1+x^{30}) \, dt \\[9pt] \implies & 1 < \int_0^1 \frac{1+x^{30}}{1+x^{60}} \, dx < 1 + \frac{1}{31} \\[9pt] \implies & \int_0^1 \frac{1+x^{30}}{1+x^{60}} \, dx = 1 + \frac{c}{31} \qquad \text{for some } 0 < c < 1. \qquad \blacksquare \end{align*}$

Find an inverse for a function defined by an integral

by RoRi

December 20, 2015

Define a function for $x \geq 0$ by

$f(x) = \int_0^x (1+t^3)^{-\frac{1}{2}} \, dt.$

Prove that $f(x)$ is strictly increasing on the nonnegative real axis.
If $g$ denotes the inverse of $f$ prove that $g''$ is proportional to $g^2$ and find the constant of proportionality.

Proof. To show $f(x)$ is strictly increasing we take the derivative,
$f(x) = \int_0^x \frac{1}{\sqrt{1+t^3}} \, dt \quad \implies \quad f'(x) = \frac{1}{\sqrt{1+x^3}}.$

Since $f'(x) > 0$ for all $x \geq 0$ we have that $f(x)$ is strictly increasing on the nonnegative real axis $. \qquad \blacksquare$
Proof. If $g$ is the inverse of $f$ then we know (Theorem 6.7 on page 252 of Apostol)
$g'(y) = \frac{1}{f'(x)}.$

But, we have defined $g$ to the function such that $g(f(x)) = x$ . Therefore,

$g'(y) = \frac{1}{f'(g(y))}.$

Therefore, we have that

$f'(x) = \frac{1}{\sqrt{1+x^3}} \quad \implies \quad g'(y) = \sqrt{1+g^3(y)}.$

Taking another derivative of $g$ with respect to $y$ we then have

$\begin{align*} g''(y) &= \frac{3 (g(y))^2 \cdot g'(y)}{2 \sqrt{1+(g(y))^3}} \\[9pt] &= \frac{3}{2} \cdot (g(y))^2 \cdot \frac{g'(y)}{\sqrt{1+(g(y))^3}} \\[9pt] &= \frac{3}{2} \cdot (g(y))^2 \cdot \frac{\sqrt{1+(g(y))^3}}{\sqrt{1+(g(y))^3}} \\[9pt] &= \frac{3}{2} (g(y))^2. \qquad \blacksquare \end{align*}$

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

by RoRi

December 20, 2015

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 $\operatorname{Li}(x)$ .

$\displaystyle{ \operatorname{Li}(x) = \frac{x}{\log x} + \int_2^x \left(\frac{dt}{\log^2 t}\right) - \frac{2}{\log 2}}$ .
$\displaystyle{ \operatorname{Li}(x) = \frac{x}{\log x} + \sum_{k=1}^{n-1} \left(\frac{k!x}{\log^{k+1} x}\right) + n! \int_2^x \left( \frac{dt}{\log^{n+1} t} \right) + C_n}$ where $C_n$ is a constant depending on $n$ . Find the value of $C_n$ for each $n$ .
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.
Let $c = 1 + \frac{1}{2} \log 2$ . Find an expression for
$\int_c^x \frac{e^{2t}}{t-1} \, dt$

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

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

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

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

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

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

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