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

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

Prove an inequality relating sums and integrals

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

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))

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.

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

  1. 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

        \[ \int_0^x f(t) \, dt = o \left( \int_0^x g(t) \, dt \right). \qquad \blacksquare\]

  2. 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.

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

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*}

Find an inverse for a function defined by an integral

Define a function for x \geq 0 by

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

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

  1. 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

  2. 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)

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*}