Home » Integration by Parts

Tag: Integration by Parts

Conclude if the given series converges or diverges and justify your conclusion

by RoRi

Test the following series for convergence or divergence. Justify the decision.

    \[ \sum_{n=1}^{\infty} \int_n^{n+1} e^{-\sqrt{x}} \, dx. \]

First, we make the substitution s = \sqrt{x}, which gives us ds = \frac{1}{2 \sqrt{x}} \, dx. Then we have

    \[ \int e^{-\sqrt{x}} \, dx = 2 \int se^{-s} \, ds. \]

This integral we evaluate using integration by parts with

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

Therefore, we have

    \begin{align*} \int e^{-\sqrt{x}} \, dx &= 2 \int se^{-s} \, ds \\[9pt] &= 2 \left( -se^{-s} + \int e^{-s} \, ds \right) \\[9pt] &= 2 (e^{-s}(-1-s)) \\ &= -2e^{-\sqrt{x}} (1+\sqrt{x}). \end{align*}

So, for the definite integral from n to n+1 we have

    \begin{align*} \int_n^{n+1} e^{-\sqrt{x}} \, dx &= -2e^{-\sqrt{n+1}} (1+\sqrt{n+1}) + 2 e^{-\sqrt{n}}(1+\sqrt{n}) \\[9pt] &= 2e^{-\sqrt{n}} (1+\sqrt{n}) - 2e^{-\sqrt{n+1}} (1+\sqrt{n+1}). \end{align*}

But then the series is a telescoping series with

    \[ b_n = 2e^{-\sqrt{n}} (1+\sqrt{n}) \quad \implies \quad \lim_{n \to \infty} b_n = 0. \]

Hence,

    \[ \sum_{n=1}^{\infty} (b_n - b_{n+1}) = b_1 - L = \frac{4}{e}. \]

Hence, the series converges.

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

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*}

Explain why ex / x cannot be integrated by parts

by RoRi

Try to use integration by parts to evaluate the integral

    \[ \int \frac{e^x}{x} \, dx. \]

If we let

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

Then we have

    \[ \int \frac{e^x}{x} \, dx = \frac{e^x}{x} + \int \frac{e^x}{x^2} \, dx. \]

If we try to integrate by parts again with u = \frac{1}{x^2} and dv = e^x then we’ll end up with another integral, this time with \frac{e^x}{x^3}. That will continue, so we’ll just keep getting integrals that we can’t evaluate.

On the other hand, if we try letting

    \begin{align*} u &= e^x & \implies && du &= e^x \, dx \\ dv &= \frac{1}{x} \, & \implies && v &= \log x, \end{align*}

Then we have

    \[ \int \frac{e^x}{x} \, dx = e^x \log x - \int e^x \log x \, dx. \]

Continuing along that route, we’ll keep getting integrals of x^k e^x \log x where k keeps getting larger.

In both cases, we keep getting increasing complicated integrals that we can never evaluate.

Evaluate the integral of (x2 + 5)1/2

by RoRi

Compute the following integral.

    \[ \int \sqrt{x^2+5} \, dx. \]

First, we want to make the substitution

    \begin{align*} x &= \sqrt{5} \tan s & \implies \qquad dx &= \sqrt{5} \sec^2 s \, ds \\ x^2 &= 5 \tan^2 s. \end{align*}

Therefore, we have

    \begin{align*} \int \sqrt{x^2+5} \, dx &= \int \sqrt{5 \tan^2 s + 5} (\sqrt{5} \sec^2 s) \, ds \\[9pt] &= 5 \int \sqrt{\tan^2 s + 1} \sec^2 s \, ds \\[9pt] &= 5 \int \sec^3 s \, ds. \end{align*}

Now, we use integration by parts, letting

    \begin{align*} u &= \sec s & \implies && du &= \sec s \tan s \, ds \\ dv &= \sec^2 s \, ds & \implies && v &= \tan s. \end{align*}

Therefore, we have

    \begin{align*} \int \sec^3 s \, ds &= \sec s \tan s - \int \tan^2 s \sec s \, ds \\[9pt] &= \sec s \tan s - \int \sec s (\sec^2 s -1) \, ds \\[9pt] &= \sec s \tan s - \int \sec^3 s \, ds + \int \sec s \, ds. \end{align*}

Moving the \int \sec^3 s \, ds back to the left side we have

    \begin{align*} && 2 \int \sec^3 s \, ds &= \sec s \tan s + \int \sec s \, ds \\[9pt] \implies && \int sec^3 s \, ds &= \frac{1}{2} \sec s \tan s + \frac{1}{2} \int \sec s \, ds \\[9pt] &&&= \frac{1}{2} \sec s \tan s + \frac{1}{2} \log |\sec s + \tan s| + C. \end{align*}

Plugging this back in above we have,

    \begin{align*} \int \sqrt{x^2+5} \, dx &= 5 \int \sec^3 s \, ds \\[9pt] &= \frac{5}{2} \sec s \tan s + \frac{5}{2} \log | \sec s + \tan s| + C \\[9pt] &= \frac{5}{2} \sec \left( \arctan \frac{x}{\sqrt{5}} \right) \frac{x}{\sqrt{5}} + \frac{5}{2} \log \left| \sec \left( \arctan \frac{x}{\sqrt{5}} \right) + \frac{x}{\sqrt{5}} \right| + C. \end{align*}

Then we note that

    \[ \sec \left( \arctan \frac{x}{\sqrt{5}} \right) = \frac{\sqrt{x^2+5}}{\sqrt{5}}. \]

(If you don’t remember your trigonometry, which I usually don’t, you can figure things like this out by drawing the triangle. Since \theta = \arctan \frac{x}{\sqrt{5}} means you have a right triangle with legs of length x and \sqrt{5}. That means the hypotenuse has length \sqrt{x^2+5}. So, \cos \theta = \frac{x}{\sqrt{x^2+5}} which then gives the above result for \sec \theta.)
Plugging this back in we then have

    \begin{align*} \int \sqrt{x^2+5} \, dx &= \frac{5}{2} \sec \left( \arctan \frac{x}{\sqrt{5}} \right) \frac{x}{\sqrt{5}} + \frac{5}{2} \log \left| \sec \left( \arctan \frac{x}{\sqrt{5}} \right) + \frac{x}{\sqrt{5}} \right| + C \\[9pt] &= \frac{x}{2} \sqrt{x^2+5} + \frac{5}{2} \log \left| \frac{x + \sqrt{x^2+5}}{\sqrt{5}} \right| + C \\[9pt] &= \frac{x}{2} \sqrt{x^2+5} + \frac{5}{2} \log \left| x + \sqrt{x^2+5} \right| - \frac{5}{2} \log \sqrt{5} + C \\[9pt] &= \frac{x}{2} \sqrt{x^2+5} + \frac{5}{2} \log \left| x + \sqrt{x^2+5} \right| + C. \end{align*}

(Where we absorbed \frac{5}{2} \log \sqrt{5} into the constant in the final line, since it is just a constant.)

Establish the integral formula for (arcsin x)/(x2)

by RoRi

Establish that the following integral formula is correct:

    \[ \int \frac{\arcsin x}{x^2} \, dx = \log \left| \frac{1-\sqrt{1-x^2}}{x} \right| - \frac{\arcsin x}{x} + C. \]

Proof. We can evaluate this integral using integration by parts. Let

    \begin{align*} u &= \arcsin x & du &= \frac{1}{\sqrt{1-x^2}} \, dx \\ dv &= \frac{1}{x^2} \, dx & v &= -\frac{1}{x}. \end{align*}

Then we have

    \[ \int \frac{\arcsin x}{x^2} \, dx &= -\frac{\arcsin x}{x} + \int \frac{1}{x \sqrt{1-x^2}} \, dx. \]

Next, we must evaluate the integral on the right. For this integral we have

    \begin{align*} \int \frac{1}{x\sqrt{1-x^2}} \, dx &= \int \frac{1 + \sqrt{1-x^2} - \sqrt{1-x^2}}{x \sqrt{1-x^2}} \, dx \\[10pt] &= \int \frac{1 + \sqrt{1-x^2}}{x \sqrt{1-x^2}} \, dx - \int \frac{1}{x} \, dx \\[10pt] &= \int \left( \frac{1+\sqrt{1-x^2}}{x \sqrt{1-x^2}} \right) \left( \frac{1-\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \right) \, dx - \log |x| \\[10pt] &= - \log |x| + \int \frac{1 - (1-x^2)}{(x \sqrt{1-x^2})(1 - \sqrt{1-x^2})} \, dx \\[10pt] &= - \log |x| + \int \left( \frac{1}{1-\sqrt{1-x^2}} \right) \left( \frac{x^2}{x\sqrt{1-x^2}} \right) \, dx \\[10pt] &= - \log |x| + \int \left( \frac{1}{1-\sqrt{1-x^2}} \right) \left( \frac{x}{\sqrt{1-x^2}} \right) \, dx. \end{align*}

Now, we make a substitution, let u = 1 - \sqrt{1-x^2}, du = \frac{x}{1-x^2} \, dx. Therefore,

    \begin{align*} \int \frac{1}{x\sqrt{1-x^2}} \, dx &= - \log |x| + \int \frac{du}{u} \\[9pt] &= - \log |x| + \log |u| + C \\[9pt] &= - \log |x| + \log \left|1 - \sqrt{1-x^2} \right| + C \\[9pt] &= \log \left| \frac{1-\sqrt{1-x^2}}{x} \right| + C. \end{align*}

Combining this with the equation above we then have

    \[ \int \frac{\arcsin x}{x^2} \, dx = \log \left| \frac{1 - \sqrt{1-x^2}}{x} \right| - \frac{\arcsin x}{x} + C. \qquad \blacksquare \]

Establish the integration formula for arcsec x

by RoRi

Establish that the following integration formula is correct:

    \[ \int \operatorname{arcsec} x \, dx = x \operatorname{arcsec} x - \frac{x}{|x|} \log \left| x + \sqrt{x^2-1} \right| + C. \]

Proof. First, we want to integrate by parts. Let

    \begin{align*} u &= \operatorname{arcsec} x & du &= \frac{1}{|x| \sqrt{x^2-1} } \, dx \\ dv &= dx & v &= x. \end{align*}

Therefore, we have

    \[ \int \operatorname{arcsec} x \, dx = x \operatorname{arcsec} x - \int \frac{x}{|x| \sqrt{x^2-1}}. \]

Now, to evaluate this integral we break it into pieces (to get rid of the |x|). Since we have

    \[ \frac{x}{|x| \sqrt{x^2-1}} = \begin{dcases} \frac{1}{\sqrt{x^2-1}} & \text{if } x > 1 \\ \frac{-1}{\sqrt{x^2-1}} & \text{if } x < -1. \end{dcases} \]

(Since \operatorname{arcsec} x is only defined for |x| \geq 1 we don’t need to worry about the cases -1 < x < 1.) We then evaluate the two pieces of the integral separately. If x > 1,

    \begin{align*} x \operatorname{arcsec} x - \int \frac{x}{|x| \sqrt{x^2-1}} \, dx &= x \operatorname{arcsec} x - \int \frac{1}{\sqrt{x^2-1}} \, dx \\ &= x \operatorname{arcsec} x - \int \frac{x + \sqrt{x^2-1}}{(\sqrt{x^2-1})(x + \sqrt{x^2-1})} \, dx \\ &= x \operatorname{arcsec} x - \int \left( \frac{1}{x+\sqrt{x^2-1}} \right) \left( \frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}} \right) \, dx \\ &= x \operatorname{arcsec} x - \int \left( \frac{1}{x+\sqrt{x^2-1}} \right) \left( 1 + \frac{x}{\sqrt{x^2-1}} \right) \, dx. \end{align*}

Now, we note that

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

Therefore, we have an integral of the form \int \frac{f'(x)}{f(x)} \, dx and so,

    \[ \int \left( \frac{1}{x+\sqrt{x^2-1}} \right) \left( \frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}} \right) \, dx = \log \left| x + \sqrt{x^2-1} \right| + C. \]

So, for x > 1 we have

    \[ \int \operatorname{arcsec} x \, dx &= x \operatorname{arcsec} x - \log \left| x + \sqrt{x^2-1} \right| + C. \]

For the case x < -1 everything is identical except we have a negative sign (since \frac{x}{|x| \sqrt{x^2-1}} has a negative sign when x < -1) so for x < -1 we have

    \[ \int \operatorname{arcsec} x \, dx = x \operatorname{arcsec} x + \log \left| x + \sqrt{x^2-1} \right| + C. \]

Therefore (since \frac{x}{|x|} = 1 if x > 1 and \frac{x}{|x|} = -1 if x < -1) we have

    \[ \int \operatorname{arcsec} x \, dx = x \operatorname{arcsec} x - \frac{x}{|x|} \log \left| x + \sqrt{x^2-1} \right| + C. \qquad \blacksquare \]