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Prove some formulas for integrals of e-t tn

by RoRi

Prove the following integral formulas.

  1. \displaystyle{ \int_0^x e^{-t} t \, dt = e^{-x} \big( e^x - 1 - x \big)}.
  2. \displaystyle{ \int_0^x e^{-t} t^2 \, dt = 2! e^{-x} \left( e^x - 1 - x - \frac{x^2}{2!} \right)}.
  3. \displaystyle{ \int_0^x e^{-t} t^3 \, dt = 3! e^{-x} \left( e^x - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!} \right)}.
  4. Guess and prove a general formula based on parts (a) – (c).
  1. Proof. We use integration by parts with

        \begin{align*} u &= t & du &= dt \\ dv &= e^{-t} \, dt & v &= -e^{-t}. \end{align*}

    This gives us

        \begin{align*} \int_0^x e^{-t} t \, dt &= -te^{-t} \Bigr \rvert_0^x + \int_0^x e^{-t} \, dt \\[9pt] &= -xe^{-x} + (-e^{-t})\Bigr \rvert_0^x \\[9pt] &= -xe^{-x} - e^{-x} + 1 \\[9pt] &= e^{-x} \left( \frac{1}{e^{-x}} - 1 - x \right) \\[9pt] &= e^{-x} \left( e^x - 1 - x \right). \qquad \blacksquare \end{align*}

  2. Proof. We use integration by parts and the result of part (a). Let

        \begin{align*} u &= t^2 & du &= 2t dt \\ dv &= e^{-t} dt & v &= -e^{-t}. \end{align*}

    This gives us

        \begin{align*} \int_0^x e^{-t} t^2 \, dt &= -t^2 e^{-t} \Bigr \rvert_0^x + 2 \int_0^x e^{-t} t \, dt \\[9pt] &= -x^2 e^{-x} + 2 \left( e^{-x} ( e^x - 1 - x ) \right) \\[9pt] &= 2! e^{-x} \left( e^x - 1 - x - \frac{x^2}{2!} \right). \qquad \blacksquare \end{align*}

  3. Proof. Again, we use integration by parts, and this time part (b). Let

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

    This gives us

        \begin{align*} \int_0^x e^{-t} t^3 \, dt &= -t^3 e^{-t} \Bigr \rvert_0^x + 3 \int_0^x e^{-t} t^2 \, dt \\[9pt] &= -x^3 e^{-x} + 3 \left( 2! e^{-x} \left( e^x - 1 - x - \frac{x^2}{2!} \right) \right) \\[9pt] &= 3! e^{-x} \left( e^x - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!} \right). \qquad \blacksquare \end{align*}

  4. Claim:

        \[ \int_0^x e^{-t} t^n \, dt = n! e^{-x} \left( e^x - \sum_{k=0}^n \frac{x^k}{k!} \right). \]

    Proof. The proof is by induction. We have already established the case n = 1 (and n = 2 and n = 3). Assume then that the formula holds for some positive integer m. We then consider the integral

        \[ \int_0^x e^{-t} t^{m+1} \, dt. \]

    Integrating by parts, we let

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

    Therefore, integrating by parts and using the induction hypothesis we have,

        \begin{align*} \int_0^x e^{-t} t^{m+1} \, dt &= -t^{m+1} e^{-t} \Bigr \rvert_0^x + (m+1) \int_0^x e^{-t} t^m \, dt \\[9pt] &= -x^{m+1}e^{-x} + (m+1) \left( m! e^{-x} \left( e^x - \sum_{k=0}^m \frac{x^k}{k!} \right) \right) \\[9pt] &= -x^{m+1} e^{-x} (m+1)! e^{-x} \left( e^x - \sum_{k=0}^m \frac{x^k}{k!} \right) \\[9pt] &= (m+1)! e^{-x} \left( e^x - \sum_{k=0}^m \frac{x^k}{k!} - \frac{x^{m+1}}{(m+1)!} \right) \\[9pt] &= (m+1)! e^{-x} \left(e^x - \sum_{k=0}^{m+1} \frac{x^k}{k!} \right). \end{align*}

    Therefore, the formula holds for m+1, and hence, for all positive integers n. \qquad \blacksquare

Compute the area and volume of solids of revolution of e-2x

by RoRi

Define the function f(x) = e^{-2x} for all x \in \mathbb{R}. Let

    \begin{align*} S(t) &= \text{the ordinate set of } f \text{ on } [0,t], \quad t> 0.\\ A(t) &= \text{the area of } S(t).\\ V(t) &= \text{the volume of the solid obtained by rotating } S(t) \text{ about the $x$-axis}.\\ W(t) &= \text{the volume of the solid obtained by rotating } S(t) \text{ about the $y$=axis}. \end{align*}

Compute

  1. A(t);
  2. V(t);
  3. W(t);
  4. \lim_{t \to 0} \frac{V(t)}{A(t)}.
  1. The area of the ordinate set on [0,t] is given by the integral,

        \[ A(t) = \int_0^t f(x) \, dx = \int_0^t e^{-2x} \, dx = -\frac{1}{2} e^{-2x} \Bigr \rvert_0^t = \frac{1}{2}(1 - e^{-2t}). \]

  2. The volume of the solid of revolution obtained by rotating f(x) about the x-axis is

        \begin{align*} V(t) &= \pi \int_0^t (f(x))^2 \, dx \\[9pt] &= \pi \int_0^t e^{-4x} \, dx \\[9pt] &= -\frac{\pi}{4} e^{-4x} \Bigr \rvert_0^t \\[9pt] &= \frac{\pi}{4} (1 - e^{-4t}). \end{align*}

  3. To compute the volume of the solid of revolution obtained by rotating f about the y-axis we first find x as a function of y.

        \[ f(x) = y = e^{-2x} \implies x = -\frac{\log y}{2}. \]

    Since f(t) = e^{-2t}, the integral is then from e^{-2t} to 1 and we have

        \begin{align*} W(t) &= \pi \int_{e^{-2t}}^1 \frac{-\log y}{2} \, dy \\[9pt] &= -\frac{\pi}{2} \int_{e^{-2t}}^1 \log y \, dy \\[9pt] &= -\frac{\pi}{2} ( y \log y - y)\Bigr \rvert_{e^{-2t}}^1 \\[9pt] &= -\frac{\pi}{2} (-1 - e^{-2t} (-2t) - e^{-2t}) \\[9pt] &= \frac{\pi}{2} (1 - e^{-2t}(2t+1)). \end{align*}

  4. Finally, using parts (c) and (d) we can compute the limit,

        \begin{align*} \lim_{t \to 0} \frac{V(t)}{A(t)} &= \lim_{t \to 0} \frac{\frac{\pi}{4} (1-e^{-4t})}{\frac{1}{2} (1-e^{-2t})} \\[9pt] &= \lim_{t \to 0} \frac{\pi (1-e^{-4t})}{2(1-e^{-2t})} \\[9pt] &= \lim_{t \to 0} \frac{\pi (e^{4t} - 1)}{2e^{2t}(e^{2t} - 1)} \\[9pt] &= \frac{\pi}{2} \lim_{t \to 0} \frac{(e^{2t}+1)(e^{2t}-1)}{e^{2t}(e^{2t}-1)} \\[9pt] &= \frac{\pi}{2} \lim_{t \to 0} \frac{e^{2t}+1}{e^{2t}} \\[9pt] &= \frac{\pi}{2} \lim_{t \to 0} \left( 1 + \frac{1}{e^{2t}} \right) \\[9pt] &= \pi. \end{align*}

Find a function given the volume of the solid of revolution it generates

by RoRi

Let f(x) be a function continuous on an interval [0,a]. The volume of the solid of revolution obtained by rotating f about the x-axis on the interval [0,a] is given by

    \[ V = a^2 + a \]

for every a > 0. Find a formula for the function f.

Using the formula for the volume of the solid of revolution generated by a function on an interval we know

    \[ V = \pi \int_0^a (f(t))^2 \, dt \quad \implies \quad a^2 + a = \pi \int_0^a (f(t))^2 \, dt. \]

Now we differentiate both sides of this equation using the fundamental theorem of calculus on the right-hand side,

    \begin{align*} x^2 + x = \pi \int_0^x (f(t))^2 \, dt && \implies && 2x + 1 = \pi (f(x))^2 \\[9pt] && \implies && f(x) = \sqrt{\frac{2x+1}{\pi}}. \end{align*}

Compute the integral from 0 to x of f(t) for the given functions

by RoRi

Find a formula to compute

    \[ F(x) = \int_0^x f(t) \, dt \]

for all x \in \mathbb{R} for the following function f(t).

  1. \displaystyle{f(t) = (t + |t|)^2}.
  2. The function,

        \[ f(t) = \begin{cases} 1-t^2 & \text{if } |t| \leq 1, \\ 1 - |t| & \text{if } |t| > 1. \end{cases} \]

  3. f(t) = e^{-|t|}.
  4. f(t) = the maximum of 1 and t^2.
  1. We know from this exercise (Section 5.5, Exercise #13) that

        \[ F(x) = \int_0^x (t + |t|)^2 \, dt = \frac{2}{3} x^2 (x + |x|). \]

  2. If \abs{x} \leq 1, then f(t) = 1-t^2 over the whole integral, and so

        \[ F(x) = \int_0^x f(t) \, dt = \int_0^x (1-t^2) \, dt = \left( t - \frac{t^3}{3} \right) \Bigr \rvert_0^x = x - \frac{x^3}{3}. \]

    Then, if x > 1 we have

        \begin{align*} F(x) &= \int_0^x f(t) \, dt \\[9pt] &= \int_0^1 (1-t^2) \, dt + \int_1^x 1-t \, dt \\[9pt] &= \left( t - \frac{t^3}{3} \right) \Bigr \rvert_0^1 + \left( t - \frac{t^2}{2} \right) \Bigr \rvert_1^x \\[9pt] &= \frac{2}{3} + x - \frac{x^2}{2} - \frac{1}{2} \\[9pt] &= x - \frac{1}{2} x|x| + \frac{|x|}{x} \frac{1}{6}. \end{align*}

    (Since x > 1 we have \frac{|x|}{x} = 1 so this equation works. This is the form Apostol wrote these answers as in the back of the book, so I’m getting our answers to match his. I wouldn’t have written them this way otherwise.)

    Finally, if x < -1 we have

        \begin{align*} F(x) &= \int_0^x f(t) \, dt \\[9pt] &= \int_0^{-1} (1-t^2) \, dt + \int_{-1}^x (1+t) \, dt \\[9pt] &= \left( t - \frac{t^3}{3} \right) \Bigr \rvert_0^{-1} + \left( t + \frac{t^2}{2} \right) \Bigr \rvert_{-1}^x \\[9pt] &= -\frac{2}{3} + x + \frac{x^2}{2} + \frac{1}{2} \\[9pt] &= x + \frac{1}{2} x|x| + \frac{|x|}{x} \frac{1}{6}. \end{align*}

    Since the formulas for F(x) are the same for x < -1 and x > 1 are the same we have

        \[ F(x) = x + \frac{1}{2} x|x| + \frac{|x|}{x} \frac{1}{6} \]

    for |x| > 1.

  3. We consider two cases. If x \geq 0 then

        \[ F(x) = \int_0^x e^{-|t|} \, dt = \int_0^x e^{-t} \, dt = -e^{-t} \Bigr \rvert_0^x = 1-e^{-x}. \]

    If x < 0 then

        \[ F(x) = \int_0^x e^{-|t|} \, dt = \int_0^x e^t \, dt = e^x - 1. \]

  4. Since the maximum of 1 and t^2 is equal to 1 if |t| \leq 1 and is equal to t^2 if |t| > 1 we consider three cases (|x| \leq 1, x > 1 and x < -1).

    For |x| \leq 1 we have

        \[ F(x) = \int_0^x f(t) \, dt = \int_0^x dt = x. \]

    For x > 1 we have

        \[ F(x) = \int_0^1 dt + \int_1^x t^2 \, dt = 1 + \frac{t^3}{3}\Bigr \rvert_1^x = \frac{x^3}{3} + \frac{2}{3} = \frac{x^3}{3} + \frac{|x|}{x} \frac{2}{3} \]

    For x < -1 we have

        \[ F(x) = \int_0^{-1} dt + \int_{-1}^x t^2 \, dt = -1 + \frac{t^3}{3} \Bigr \rvert_{-1}^x = +\frac{x^3}{3} + \frac{|x|}{x} \frac{2}{3}. \]

Prove an identity of given finite sums

by RoRi

Prove the identity:

    \[ \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{k+m+1} = \sum_{k=0}^m (-1)^k \binom{m}{k} \frac{1}{k+n+1}. \]

Proof. Using the hint (that \frac{1}{k+m+1} = \int_0^1 t^{k+m} \, dt) we start with the expression on the left,

    \begin{align*} \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{k+m+1} &= \sum_{k=0}^n (-1)^k \binom{n}{k} \int_0^1 t^{m+k} \, dt \\[10pt] &= \sum_{k=0}^n \int_0^1 (-1)^k \binom{n}{k} t^m t^k \, dt \\[10pt] &= \int_0^1 \sum_{k=0}^n (-1)^k \binom{n}{k} t^m t^k \, dt &(\text{finite sum}) \\[10pt] &= \int_0^1 t^m \sum_{k=0}^n \binom{n}{k} (-t)^k \, dt \\[10pt] &= \int_0^1 t^m \sum_{k=0}^n \binom{n}{k} (-t)^k (1)^{n-k} \, dt \\[10pt] &= \int_0^1 t^m (1-t)^n \, dt &(\text{Binomial theorem}). \end{align*}

(The interchange of the sum and integral is fine since it is a finite sum. Those planning to take analysis should note that this cannot always be done in the case of infinite sums.) Now, we have a reasonable integral, but we still want to get everything back into the form of the sum on the right so we make the substitution s = 1-t, ds = -dt. This gives us new limits of integration from 1 to 0. Therefore, we have

    \begin{align*} \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{k+m+1} &= \int_0^1 t^m (1-t)^n \, dt \\[10pt] &= -\int_1^0 (1-s)^m s^n \, ds \\[10pt] &= \int_0^1 (1-s)^m s^n \, ds \\[10pt] &= \int_0^1 s^n \sum_{k=0}^m \binom{m}{k} (-s)^k 1^{m-k} \, ds &(\text{Binomial theorem})\\[10pt] &= \int_0^1 \sum_{k=0}^m \binom{m}{k} (-1)^k s^{k+n} \, ds \\[10pt] &= \sum_{k=0}^m (-1)^k \binom{m}{k} \int_0^1 s^{k+n} \, ds \\[10pt] &= \sum_{k=0}^m (-1)^k \binom{m}{k} \frac{1}{k+n+1}. \qquad \blacksquare \end{align*}

Evaluate given integrals in terms of the integral from 0 to 1 of et / (t+1)

by RoRi

Define an integral function

    \[ A = \int_0^1 \frac{e^t}{t-a-1} \, dt. \]

In terms of A evaluate the following:

  1. \displaystyle{ \int_{a-1}^a \frac{e^{-t}}{t-a-1} \, dt}.
  2. \displaystyle{ \int_0^1 \frac{te^{t^2}}{t^2+1} \, dt}.
  3. \displaystyle{ \int_0^1 \frac{e^t}{(t+1)^2} \, dt}.
  4. \displaystyle{ \int_0^1 e^t \log (1+t) \, dt}.
  1. First, we make the substitution s = -t+a so ds = -dt. The bounds of integration are then

        \[ -(a-1) + a = 1 \qquad \text{and} \qquad -a + a = 0. \]

    Therefore we have,

        \begin{align*} \int_{a-1}^a \frac{e^{-t}}{t-a-1} \, dt &= -\int_1^0 \frac{e^{s-a}}{-s-1} \, ds \\ &= e^{-a} \int_1^0 \frac{e^s}{s+1} \, ds \\ &= -e^{-a} \int_0^1 \frac{e^s}{s+1} \, ds \\ &= -Ae^{-a}. \end{align*}

  2. For this one, make the substitution s = t^2, ds = 2t \, dt. The bounds of integration don’t change since 0^2 = 0 and 1^2 = 1. So we have,

        \begin{align*} \int_0^1 \frac{te^{t^2}}{t^2+1} \, dt &= \frac{1}{2} \int_0^1 \frac{2te^{t^2}}{t^2+1} \, dt \\ &= \frac{1}{2} \int_0^1 \frac{e^s}{s+1} \, ds \\ &= \frac{1}{2} A. \end{align*}

  3. To compute this in terms of A, we integrate by parts. Let

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

    Therefore we have

        \begin{align*} \int_0^1 \frac{e^t}{(t+1)^2} \, dt &= -\frac{e^t}{t+1} \Bigr \rvert_0^1 + \int_0^1 \frac{e^t}{t+1} \, dt \\ &= -\frac{e}{2} + 1 + A \\ &= A + 1 - \frac{1}{2}e. \end{align*}

  4. We use integration by parts again, this time let

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

    Therefore we have,

        \begin{align*} \int_0^1 e^t \log(1+t) \, dt &= e^t \log(1+t) \Bigr \rvert_0^1 - \int_0^1 \frac{e^t}{t+1} \, dt \\ &= e \log 2 - A. \end{align*}

Conjecture and prove a statement about a differentiable function satisfying f(x+a) = bf(x)

by RoRi

Let f(x) be a function which is differentiable everywhere and which satisfies

    \[ f(x+a) = bf(x) \]

for some positive constants a and b. What can you conclude about such a function f?

(Note: I’m not entirely sure what Apostol wants here since the instruction “what can you conclude” is pretty vague. He does give an “answer” in the back of the book, so I verify that it does have the properties indicated, but I don’t know how you would arrive at that expression just from the question statement. I’ll mark this question as incompletely and hopefully come up with something better in the future.)

Since f(x) satisfies the functional equation f(x+a) = bf(x) we can write

    \[ f(x) = b^{\frac{x}{a}} g(x) \]

which implies

    \[ g(x) = \left( \frac{1}{b} \right)^{\frac{x}{a}} f(x). \]

Then computing

    \begin{align*} g(x+a) &= \left( \frac{1}{b} \right)^{\frac{x+a}{a}} f(x+a) \\ &= \left(\frac{1}{b}\right)^{\frac{x}{a}} \cdot \frac{1}{b} \cdot bf(x) \\ &= \left( \frac{1}{b} \right)^{\frac{x}{a}} f(x) \\ &= g(x). \end{align*}

Thus, g(x) is indeed periodic with period a and so

    \begin{align*} f(x+a) &= b^{\frac{x+a}{a}} g(x+a) \\ &= b\cdot b^{\frac{x}{a}} g(x) \\ &= b f(x). \end{align*}

So, this definition of f(x) in terms of the periodic function g(x) indeed satisfies the functional equation.

Find continuous functions satisfying given conditions

by RoRi

Find continuous functions which satisfy the given conditions for all x \in \mathbb{R}.

  1. \displaystyle{ \int_0^x f(t) \, dt = e^x}.
  2. \displaystyle{ \int_0^{x^2} f(t) \, dt = 1 - 2^{x^2}}.
  3. \displaystyle{ \int_0^x f(t) \, dt = f^2(x) - 1}.
  1. No such function can exist since for x = 0 we have

        \[ \int_0^x f(t) \, dt = \int_0^0 f(t) \, dt = 0 \neq e^0 = 1. \]

  2. Taking derivatives of both sides of the given equation we have

        \begin{align*} &&\frac{d}{dx} \left(\int_0^{x^2} f(t) \, dt\right) &= \frac{d}{dx} \left( 1 - 2^{(x^2)} \right) \\[9pt] \implies && (2x)(f(x^2)) &= \frac{d}{dx} \left(1 - e^{x^2 \log 2} \right) \\[9pt] \implies && (2x)(f(x^2)) &= -(2x \log 2) e^{x^2 \log 2} \\[9pt] \implies && f(x^2) &= -(\log 2)(2^{(x^2)}) \\[9pt] \implies && f(x) &= - 2^x \log 2. \end{align*}

  3. Again, taking derivatives of both sides we have

        \begin{align*} &&\frac{d}{dx} \left( \int_0^x f(t) \, dt \right) &= \frac{d}{dx} \left( f^2(x) - 1 \right) \\[9pt] \impies && f(x) &= 2f(x)f'(x) \\[9pt] \implies && 1 &= 2f'(x) \end{align*}

    at all points x such that f(x) \neq 0. (Since \int_0^x f(t) \, dt = f^2(x) - 1 is not satisfied by the zero function f(x) = 0, we know there are real x such that f(x) \neq 0.) Then, integrating

        \[ f'(x) = \frac{1}{2} \quad \implies \quad f(x) = \frac{1}{2} x + C. \]

    Now, we can solve for C by evaluating the given identity at x = 0,

        \begin{align*} && \int_0^x f(t) \, dt &= f^2 (x) - 1 \\[9pt] \implies && \int_0^0 f(t) \, dt &= f^2 (0) - 1 \\[9pt] \implies && 0 &= \left( \frac{0}{2} + C\right)^2 - 1\\[9pt] \implies && 0 &= C^2 - 1 \\[9pt] \implies && C &= \pm 1. \end{align*}

    Therefore, we have

        \[ f(x) = \frac{1}{2} x \pm 1. \]

Determine some properties of the integral of et / t

by RoRi

Let

    \[ F(x) = \int_1^x \frac{e^t}{t} \, dt \qquad \text{for} \quad x > 0. \]

  1. Find all values of x > 0 such that log x \leq F(x).
  2. Prove that

        \[ \int_1^x \frac{e^t}{t+a} \, dt = e^{-a} (F(x+a) - F(1+a)). \]

  3. Similar to part (b), find expressions for

        \[ \int_1^x \frac{e^{at}}{t} \, dt, \qquad \int_1^x \frac{e^t}{t^2} \, dt, \qquad \int_1^x \e^{\frac{1}{t}} \, dt. \]

  1. Recalling the definition of \log x as the integral,

        \[ \log x = \int_1^x \frac{1}{t} \, dt, \]

    We compute,

        \[ F(x) - \log x = \int_1^x \frac{e^t}{t} \, dt - \int_1^x \frac{1}{t} \, dt = \int_1^x \frac{e^t -1}{t} \, dt. \]

    Then, since

        \[ \frac{e^t - 1}{t} \geq 0 \qquad \text{for all } t > 0, \]

    we have

        \[ \int_1^x \frac{e^t - 1}{t} \, dt \geq 0 \qquad \text{if} \quad x \geq 1, \]

    and

        \[ \int_1^x \frac{e^t - 1}{t} \, dt = - \int_x^1 \frac{e^t - 1}{t} \, dt < 0 \qquad \text{if} \quad x < 1. \]

    Hence, the set of x such that \log x \leq F(x) is all real x \geq 1.

  2. Proof. We start by computing,

        \begin{align*} \int_1^x \frac{e^t}{t+a} &= \frac{1}{e^a} \int_1^x \frac{e^t e^a}{t+a} \, dt \\[9pt] &= e^{-a} \int_1^x \frac{e^{t+a}}{t+a} \, dt. \end{align*}

    Then, using the translation invariance of the integral,

        \begin{align*} e^{-a} \int_1^x \frac{e^{t+a}}{t+a} \, dt &= e^{-a} \int_{1+a}^{x+a} \frac{e^{(t-a)+a}}{(t-a)+a} \, dt \\[9pt] &= e^{-a} \int_{1+a}^{x+a} \frac{e^t}{t} \, dt \\[9pt] &= e^{-a} \left( \int_1^{x+a} \frac{e^t}{t} \, dt - \int_1^{1+a} \frac{e^t}{t} \, dt \right) \\[9pt] &= e^{-a} (F(x+a) - F(1+a)). \qquad \blacksquare \end{align*}

  3. We claim the following formulas hold:

        \begin{align*} \int_1^x \frac{e^{at}}{t} \, dt &= F(ax) - F(a) \\[10pt] \int_1^x \frac{e^t}{t^2} \, dt &= F(x) - \frac{e^x}{x} + e \\[10pt] \int_1^x e^{\frac{1}{t}} \, dt &= xe^{frac{1}{x}} - e - F \left( \frac{1}{x} \right). \end{align*}

    Proof. Making the substitution s = at, ds = a \, dt we have

        \begin{align*} \int_1^x \frac{e^{at}}{t} \, dt &= \int_a^{ax} \frac{e^s}{\frac{s}{a}} \frac{1}{a} \, ds \\[9pt] &= \int_a^{ax} \frac{e^s}{s} \, ds \\[9pt] &= \int_1^{ax} \frac{e^s}{s} \, ds - \int_1^a \frac{e^s}{s} \, ds \\[9pt] &= F(ax) - F(a). \qquad \blacksquare \end{align*}

    Proof. To prove this identity we integrate by parts, letting

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

    Therefore,

        \begin{align*} \int_1^x \frac{e^t}{t^2} \, dt &= \left. -\frac{e^t}{t}\right|_1^x + \int_1^x \frac{e^t}{t} \, dt \\[9pt] &= e - \frac{e^x}{x} + F(x) \\[9pt] &= F(x) - \frac{e^x}{x} + e. \qquad \blacksquare \end{align*}

    Proof. For the final identity, we use the substitution s = \frac{1}{t}, ds = -\frac{1}{t^2} \, dt. This gives us

        \[ \int_1^x e^{\frac{1}{t}} \, dt &= -\int_1^{\frac{1}{x}} \frac{e^s}{s^2} \, ds. \]

    Now, we use integration by parts with

        \begin{align*} u &= e^s & \implies && du &= e^s \, ds \\ dv &= \frac{1}{s^2} \, ds & \implies && v &= -\frac{1}{s}. \end{align*}

    This gives us

        \begin{align*} \int_1^x e^{\frac{1}{t}} \, dt &= - \int_1^{\frac{1}{x}} \frac{e^s}{s^2} \, ds \\[9pt] &= \frac{e^s}{s} \Bigr \rvert_1^{\frac{1}{x}} - \int_1^{\frac{1}{x}} \frac{e^s}{s} \, ds \\[9pt] &= xe^{\frac{1}{x}} - e - F \left( \frac{1}{x} \right). \qquad \blacksquare \end{align*}