6.1 Integration by Parts

In physics and engineering the Gamma function^[1] $\Gamma\,(t)$, defined by

\[ \Gamma\,(t) ~=~ \int_0^{\infty} x^{t-1} \, e^{-x} ~\dx \quad\text{for all $t > 0$,} \]

has found many uses. Evaluating $\Gamma\,(2)$ entails integrating the function $f(x) = x\,e^{-x}$. No formula or substitution you have learned so far would be of help. Differentiating that function, on the other hand, is easy. By the Product Rule for differentials,

\begin{align*} d(x\,e^{-x}) ~&=~ x\,d(e^{-x}) ~+~ d(x)\,e^{-x}\\ &=~ -x\,e^{-x}\,\dx ~+~ e^{-x}\,\dx\\ d(x\,e^{-x}) ~&=~ -x\,e^{-x}\,\dx ~-~ d(e^{-x})\\ x\,e^{-x}\,\dx ~&=~ -d(x\,e^{-x}) ~-~ d(e^{-x}),\quad\text{so integrate both sides to get}\\[6pt] \int x\,e^{-x}\,\dx ~&=~ -\int d(x\,e^{-x}) ~-~ \int d(e^{-x}) ~=~ -x\,e^{-x} ~-~ e^{-x} ~+~ C \end{align*}

since $\int d\!F \,=\, F \,+\, C$. Generalizing this process for functions $u$ and $v$,

\begin{align*} d(uv) ~&=~ u\,\dv ~+~ v\,\du\\ u\,\dv ~&=~ d(uv) ~-~ v\,\du \end{align*}

so that integrating both sides yields the integration by parts formula:

Theorem 6.1

For differentiable functions $u$ and $v$:

\begin{equation} \int u\,\dv ~=~ uv ~-~ \int v\,\du \tag{6.1}\end{equation}

Integration by parts is just the Product Rule for derivatives in integral form, typically used when the integral $\int v\,\du$ would be simpler than the original integral $\int u\,\dv$.

Example 6.1

Use integration by parts to evaluate $~\displaystyle\int x\,e^{-x}\,\dx~$. Use the answer to evaluate $\Gamma\,(2)$.

Solution: The original integral is always of the form $\int u\,\dv$, so you must decide which parts of $x e^{-x}\,\dx$ will represent $u$ and $\dv$. Typically you would choose $\dv$ to be a differential that you could integrate easily (since you will need to integrate $\dv$ to get $v$) and choose $u$ to be a function whose derivative is simpler than $u$ (since that derivative will appear in $v\,\du$, which you hope to be a simpler integral). In this case, pick $u = x$ and $\dv = e^{-x}\,\dx$. Then $\du = \dx$ and $v = \int \dv = \int e^{-x}\,\dx = -e^{-x}$ (you can omit the generic constant $C$ for now—include it when you have evaluated $\int v\,\du$). Thus,

\begin{align*} \int u\,\dv ~&=~ uv ~-~ \int v\,\du\\[6pt] \int \underbracket[0.3pt]{x\vphantom{\tfrac{1}{2}}}_{u\vphantom{d}}\,\underbracket[0.3pt]{e^{-x}\,\dx\vphantom{\tfrac{1}{2}}}_{\dv} ~&=~ \underbracket[0.3pt]{x\vphantom{\tfrac{1}{2}}}_{u\vphantom{d}}\,\underbracket[0.3pt]{(-e^{-x})\vphantom{\tfrac{1}{2}}}_{v\vphantom{d}} ~-~ \int \underbracket[0.3pt]{-e^{-x}\vphantom{\tfrac{1}{2}}}_{v\vphantom{d}}\,\underbracket[0.3pt]{\dx\vphantom{\tfrac{1}{2}}}_{\du}\\[6pt] \int x\,e^{-x}\,\dx ~&=~ -x\,e^{-x} ~-~ e^{-x} ~+~ C \end{align*}

which agrees with the example at the beginning of this section. Note that $\int v\,\du = \int -e^{-x}\,\dx$, which indeed is simpler than the original integral. The Gamma function value $\Gamma\,(2)$ can now be evaluated:

\begin{align*} \Gamma\,(2) ~&=~ \int_0^{\infty} x\,e^{-x}\,\dx\\ &=~ -x\,e^{-x} ~-~ e^{-x}~\Biggr|_{0}^{\infty}\\[4pt] &=~ \lim_{x \to \infty}~(-x\,e^{-x} ~-~ e^{-x}) ~-~ (-0\,e^{0} ~-~ e^{0})\\[4pt] &=~ -\left(\lim_{x \to \infty}~\frac{x}{e^x}\right) ~-~ \left(\lim_{x \to \infty}~e^{-x}\right) ~-~ (0 - 1)\\[4pt] &=~ 0 ~-~ 0 ~+~ 1 ~=~ 1 \end{align*}

Example 6.2

What would happen in Example 6.1 if you let $u = e^{-x}$ and $\dv = x\,\dx$?

Solution: In this case $\du = -e^{-x}\,\dx$ and $v = \int \dv = \int x\,\dx = \frac{1}{2}x^2$, so that

\begin{align*} \int x\,e^{-x}\,\dx ~&=~ \int \underbracket[0.3pt]{e^{-x}\vphantom{\tfrac{1}{2}}}_{u\vphantom{d}}\,\underbracket[0.3pt]{x~\dx\vphantom{\tfrac{1}{2}}}_{\dv}\\[6pt] &=~ \underbracket[0.3pt]{e^{-x}\vphantom{\frac{1}{2}}}_{u\vphantom{d}}\,\underbracket[0.3pt]{\frac{1}{2}x^2}_{v\vphantom{d}} ~-~ \int \underbracket[0.3pt]{\frac{1}{2}x^2}_{v\vphantom{d}}\,\underbracket[0.3pt]{(-e^{-x})\,\dx\vphantom{\frac{1}{2}}}_{\du}\\[6pt] &=~ \frac{1}{2}x^2\,e^{-x} ~+~ \frac{1}{2}\,\int x^2\,e^{-x}\,\dx \end{align*}

which leads you in the wrong direction: a more difficult integral than the original.

Example 6.2 showed the importance of an appropriate choice for $u$ and $\dv$. There are some rough guidelines for that choice—as in Example 6.1—but no rules that are guaranteed to always work. It might not be clear when you should even attempt integration by parts.

Example 6.3

Evaluate $~\displaystyle\int \ln x~\dx~$.

Solution: Integration by parts ostensibly requires two functions in the integral, whereas here $\ln x$ appears to be the only one. However, the choice for $\dv$ is a differential, and one exists here: $\dx$. Choosing $\dv = \dx$ obliges you to let $u = \ln\,x$. Then $\du = \frac{1}{x}\;\dx$ and $v = \int \dv = \int \dx = x$. Now integrate by parts:

\begin{align*} \int u\,\dv ~&=~ uv ~-~ \int v\,\du\\[6pt] \int \ln\,x~\dx ~&=~ (\ln\,x)\,(x) ~-~ \int x \cdot \frac{1}{x}~\dx\\[6pt] &=~ x\,\ln\,x ~-~ \int 1~\dx\\[6pt] &=~ x\,\ln\,x ~-~ x ~+~ C \end{align*}

Note that choosing $\dv = \ln\,x\;\dx$ would be pointless, as integrating $\dv$ to get $v$ is the original problem!

Example 6.4

Evaluate $~\displaystyle\int x^3\,e^{x^2}\,\dx~$.

Solution: One frequently useful guideline for integration by parts is to eliminate the most complicated function in the integral by integrating it—as $\dv$—into something simpler (which becomes $v$). In this integral, $e^{x^2}$ is somewhat complicated but has no closed form antiderivative. However, $x\,e^{x^2}$ appears in the integral and can be integrated easily (using a substitution as in Section 5.4). So choose $\dv = x\,e^{x^2}\,\dx$, which means $u = x^2$. Then $\du = 2x\,\dx$ and $v = \int \dv = \int x\,e^{x^2}\,\dx = \frac{1}{2}e^{x^2}$. Now integrate by parts:

\begin{align*} \int u\,\dv ~&=~ uv ~-~ \int v\,\du\\[6pt] \int x^3\,e^{x^2}\,\dx ~&=~ x^2\,\cdot\,\frac{1}{2}\,e^{x^2} ~-~ \int \frac{1}{2}\,e^{x^2} \cdot 2x~\dx\\[6pt] &=~ \frac{x^2}{2}\,e^{x^2} ~-~ \int x\,e^{x^2}\,\dx\\[6pt] &=~ \frac{x^2}{2}\,e^{x^2} ~-~ \frac{1}{2}\,e^{x^2} ~+~ C\\[6pt] &=~ \frac{(x^2 - 1)}{2}\,e^{x^2} ~+~ C \end{align*}

Sometimes multiple rounds of integration by parts are needed, as in the following example.

Example 6.5

Evaluate $~\displaystyle\int x^2\,e^{-x}\,\dx~$.

Solution: This integral appears similar to the one in Example 6.1, so choose $\dv = e^{-x}\,\dx$ and $u = x^2$. Then $\du = 2x\,\dx$ and $v = \int e^{-x}\,\dv = -e^{-x}$. Now integrate by parts:

\begin{align*} \int u\,\dv ~&=~ uv ~-~ \int v\,\du\\[6pt] \int x^2\,e^{-x}\,\dx ~&=~ x^2\,\cdot\,(-e^{-x}) ~-~ \int -e^{-x} \cdot 2x~\dx\\[6pt] &=~ -x^2\,e^{-x} ~+~ 2\,\int x\,e^{-x}\,\dx\quad\text{(integrate by parts \emph{again})}\\[4pt] \int x^2\,e^{-x}\,\dx ~&=~ -x^2\,e^{-x} ~+~ 2\,(-x\,e^{-x} ~-~ e^{-x}) ~+~ C\quad\text{(by Example 6.1)}\\ &=~ -x^2\,e^{-x} ~-~ 2x\,e^{-x} ~-~ 2\,e^{-x} ~+~ C \end{align*}

In the above example, notice that the $u=2x$ in the second integral came from the derivative of the $u=x^2$ in the first integral. Likewise, the $\dv =-e^{-x}\,\dx$ in the second integral came from integrating the $\dv=e^{-x}\,\dx$ in the first integral. In general, if $n$ rounds of integration by parts were needed, with $u_i$ and $v_i$ representing the $u$ and $v$, respectively, for round $i =1$, $2$, $\ldots$, $n$, then the repeated integration by parts would look like this:

\begin{align*} \int u_1\,\dv_1 ~&=~ u_1v_1 ~-~ \int v_1\,\du_1\\[6pt] &=~ u_1v_1 ~-~ \int u_2\,\dv_2\\[6pt] &=~ u_1v_1 ~-~ \left(u_2v_2 ~-~ \int v_2\,\du_2\right) ~=~ u_1v_1 ~-~ u_2v_2 ~+~ \int u_3\,\dv_3\\[6pt] &=~ u_1v_1 ~-~ u_2v_2 ~+~ \left(u_3v_3 ~-~ \int u_4\,\dv_4\right)\\[6pt] &=~ u_1v_1 ~-~ u_2v_2 ~+~ u_3v_3 ~-~ \left(u_4v_4 ~-~ \int u_5\,\dv_5\right)\\[6pt] &=~ \cdots\\[6pt] &=~ u_1v_1 ~-~ u_2v_2 ~+~ u_3v_3 ~-~ u_4v_4 ~+~ u_5v_5 ~-~ \cdots ~ \int u_n\,\dv_n \end{align*}

The last integral $\int u_n\,\dv_n$ is one you could presumably integrate easily. The above procedure is called the tabular method for integration by parts, since it can be shown in a table (the arrows indicate multiplication):

$tikzpicture[>=latex] [left] at (0,0.2) u; [right] at (1,0.2) ; (-0.6,-0.15) -- (1.7,-0.15); [left] at (0,-0.5) u_1; [right] at (1,-0.5) _1; [left] at (0,-1) u_2; [right] at (1,-1) v_1; [->] (0,-0.5) -- (1,-1); [right] at (2,-1) (+); [->] (3,-1) -- (4,-1) node[right] + u_1\,v_1; [left] at (0,-1.5) u_3; [right] at (1,-1.5) v_2; [->] (0,-1) -- (1,-1.5); [right] at (2,-1.5) (--); [->] (3,-1.5) -- (4,-1.5) node[right] - u_2\,v_2; [left] at (0,-2) u_3; [right] at (1,-2) v_3; [->] (0,-1.5) -- (1,-2); [right] at (2,-2) (+); [->] (3,-2) -- (4,-2) node[right] + u_3\,v_3; [left] at (0,-2.5) u_4; [right] at (1,-2.5) v_4; [->] (0,-2) -- (1,-2.5); [right] at (2,-2.5) (--); [->] (3,-2.5) -- (4,-2.5) node[right] - u_4\,v_4; [left] at (0,-3) ; [right] at (1,-3) ; [right] at (2,-3) ; tikzpicture$

The idea is to differentiate down the $u$ column and integrate down the $\dv$ column. If the $u$ in the original integral is a polynomial of degree $n$, then you know from Section 1.6 that its $(n+1)$-st derivative will be 0, at which point the tabular method terminates. The integral is then the sum of the indicated products with alternating signs.

For example, the tabular method on the integral from Example 6.5 looks like this:

$tikzpicture[>=latex] [left] at (0,0.2) u; [right] at (1,0.2) ; (-0.6,-0.15) -- (2.3,-0.15); [left] at (0,-0.5) x^2; [right] at (1,-0.5) e^-x\, ; [left] at (0,-1) 2x; [right] at (1,-1) -e^-x; [->] (0,-0.5) -- (1,-1); [right] at (2.4,-1) (+); [->] (3.4,-1) -- (4.4,-1) node[right] + (x^2)\,(-e^-x); [left] at (0,-1.5) 2; [right] at (1,-1.5) e^-x; [->] (0,-1) -- (1,-1.5); [right] at (2.4,-1.5) (--); [->] (3.4,-1.5) -- (4.4,-1.5) node[right] - (2x)\,(e^-x); [left] at (0,-2) 0; [right] at (1,-2) -e^-x; [->] (0,-1.5) -- (1,-2); [right] at (2.4,-2) (+); [->] (3.4,-2) -- (4.4,-2) node[right] + (2)\,(-e^-x); [->] (-1.5,-2) node[left] STOP -- (-0.5,-2); tikzpicture$

The integral is the sum of the products, and agrees with the result in Example 6.5:

\[ \int x^2\,e^{-x}\,\dx ~=~ +~(x^2)\,(-e^{-x}) ~-~(2x)\,(e^{-x}) ~+~(2)\,(-e^{-x}) ~+~ C ~=~ -x^2\,e^{-x} ~-~ 2x\,e^{-x} ~-~ 2\,e^{-x} ~+~ C \]

Example 6.6

Evaluate $~\displaystyle\int x^3\,e^{-x}\,\dx~$.

Solution: Use the tabular method with $u = x^3$ and $\dv = e^{-x}\,\dx$:

$tikzpicture[>=latex,every node/.style=font=] [left] at (0,0.2) u; [right] at (1,0.2) ; (-0.6,-0.15) -- (2.1,-0.15); [left] at (0,-0.5) x^3; [right] at (1,-0.5) e^-x\, ; [left] at (0,-1) 3x^2; [right] at (1,-1) -e^-x; [->] (0,-0.5) -- (1,-1); [right] at (2.2,-1) (+); [->] (3.1,-1) -- (4.1,-1) node[right] + (x^3)\,(-e^-x); [left] at (0,-1.5) 6x; [right] at (1,-1.5) e^-x; [->] (0,-1) -- (1,-1.5); [right] at (2.2,-1.5) (--); [->] (3.1,-1.5) -- (4.1,-1.5) node[right] - (3x^2)\,(e^-x); [left] at (0,-2) 6; [right] at (1,-2) -e^-x; [->] (0,-1.5) -- (1,-2); [right] at (2.2,-2) (+); [->] (3.1,-2) -- (4.1,-2) node[right] + (6x)\,(-e^-x); [left] at (0,-2.5) 0; [right] at (1,-2.5) e^-x; [->] (0,-2) -- (1,-2.5); [right] at (2.2,-2.5) (--); [->] (3.1,-2.5) -- (4.1,-2.5) node[right] - (6)\,(e^-x); [->] (-1.5,-2.5) node[left] STOP -- (-0.5,-2.5); tikzpicture$

\[ \int x^3\,e^{-x}\,\dx ~=~ -x^3\,e^{-x} ~-~ 3x^2\,e^{-x} ~-~ 6x\,e^{-x} ~-~ 6\,e^{-x} ~+~ C \]

Integration by parts can sometimes result in the original integral reappearing, allowing it to be combined with the original integral.

Example 6.7

Evaluate $~\displaystyle\int \sec^3 x~\dx~$.

Solution: Let $u=\sec\,x$ and $\dv=\sec^2 x\;\dx$, so that $\du = \sec\,x\;\tan\,x\;\dx$ and $v = \int \dv = \int \sec^2 x\,\dx = \tan\,x$. Then

\begin{align*} \int u\,\dv ~&=~ uv ~-~ \int v\,\du\\[6pt] \int \sec^3 x~\dx ~&=~ \sec\,x\;\tan\,x ~-~ \int \sec\,x\;\tan^2 x~\dx\\[6pt] \int \sec^3 x~\dx ~&=~ \sec\,x\;\tan\,x ~-~ \int \sec\,x\;(\sec^2 x \,-\, 1)~\dx\\[6pt] \int \sec^3 x~\dx ~&=~ \sec\,x\;\tan\,x ~+~ \int \sec\,x~\dx ~-~ \int \sec^3 x~\dx\\[6pt] 2\,\int \sec^3 x~\dx ~&=~ \sec\,x\;\tan\,x ~+~ \ln\;\abs{\,\sec\,x \;+\; \tan\,x\,} ~+~ C\\[6pt] \int \sec^3 x~\dx ~&=~ \frac{1}{2}\,\left(\sec\,x\;\tan\,x ~+~ \ln\;\abs{\,\sec\,x \;+\; \tan\,x\,}\right) ~+~ C \end{align*}

Example 6.8

Evaluate $~\displaystyle\int e^x\,\sin\,x~\dx~$.

Solution: Let $u=e^x$ and $\dv=\sin\,x\,\dx$, so that $\du = e^x\,\dx$ and $v = \int \dv = \int \sin\,x\,\dx = -\cos\,x$. Then

\begin{align*} \int u\,\dv ~&=~ uv ~-~ \int v\,\du\\[6pt] \int e^x\,\sin\,x~\dx ~&=~ -e^x\,\cos\,x ~+~ \int e^x\,\cos\,x~\dx \end{align*}

and so integration by parts is needed again, for the integral on the right: let $u=e^x$ and $\dv=\cos\,x\,\dx$, so that $\du = e^x\,\dx$ and $v = \int \dv = \int \cos\,x\,\dx = \sin\,x$. Then

\begin{align*} \int e^x\,\sin\,x~\dx ~&=~ -e^x\,\cos\,x ~+~ \left(uv ~-~ \int v\,\du\right)\\[6pt] \int e^x\,\sin\,x~\dx ~&=~ -e^x\,\cos\,x ~+~ \left(e^x\,\sin\,x ~-~ \int e^x\,\sin\,x~\dx\right)\\[6pt] 2\,\int e^x\,\sin\,x~\dx ~&=~ -e^x\,\cos\,x ~+~ e^x\,\sin\,x\\[6pt] \int e^x\,\sin\,x~\dx ~&=~ \frac{e^x}{2}\,(\sin\,x ~-~ \cos\,x) ~+~ C \end{align*}

In Example 6.1 integration by parts was used in evaluating an improper integral. In general, in definite or improper integrals where $a$ and $b$ are real numbers or $\pm\,\infty$,

\[ \int_a^b u\,\dv ~=~ uv~\Biggr|_a^b ~-~ \int_a^b v\,\du ~. \]

Example 6.9

Evaluate $~\displaystyle\int_0^1 x^3\,\sqrt{1-x^2}~\dx~$.

Solution: Since $x^3\,\sqrt{1-x^2} =x^2\,\cdot\,x\,\sqrt{1-x^2}$, and $x\,\sqrt{1-x^2}$ is easy to integrate (via a substitution), let $u=x^2$ and $\dv=x\,\sqrt{1-x^2}\,\dx$. Then $\du = 2x\,\dx$ and $v = \int \dv = \int x\,\sqrt{1-x^2}\,\dx = -\frac{1}{3}(1-x^2)^{3/2}$, and so:

\begin{align*} \int_a^b u\,\dv ~&=~ uv~\Biggr|_a^b ~-~ \int_a^b v\,\du\\[6pt] \int_0^1 x^3\,\sqrt{1-x^2}~\dx ~&=~ -\frac{x^2}{3}(1-x^2)^{3/2}~\Biggr|_0^1 ~+~ \int_0^1 \frac{2x}{3}(1-x^2)^{3/2}~\dx\\[6pt] &=~ (0 - 0) ~+~ \left(\frac{-2}{15}(1-x^2)^{5/2}~\Biggr|_0^1\right) ~=~ 0 ~+~ \frac{2}{15} ~=~ \frac{2}{15} \end{align*}

Exercises

For Exercises 1-25, evaluate the given integral.

$\displaystyle\int x\,\ln\,x~\dx$
$\displaystyle\int x^2\,e^{x}\,\dx$
$\displaystyle\int x\,\cos\,x~\dx$
$\displaystyle\int x\,3^x\,\dx$
$\displaystyle\int x^2\,a^x\,\dx~~(a>0)$

5
$\displaystyle\int \ln\,4x~\dx$
$\displaystyle\int \ln\,x^2~\dx$
$\displaystyle\int x^2\,\sin\,x~\dx$
$\displaystyle\int x\,\cos^2\,x~\dx$
$\displaystyle\int \sin\,x\;\cos\,2x~\dx$

5
$\displaystyle\int \sin^{-1} x~\dx$
$\displaystyle\int \cos^{-1}\,2x~\dx$
$\displaystyle\int \tan^{-1}\,3x~\dx$
$\displaystyle\int x\,\sec^2 x~\dx$
$\displaystyle\int \sin\,x\;\sin\,3x~\dx$

5
$\displaystyle\int \frac{\ln\,x}{x^3}\,\dx\vphantom{\displaystyle\int_0^2 \frac{x^3\,\dx}{\sqrt{4 - x^2}}}$
$\displaystyle\int x^3\,\ln^2 x~\dx\vphantom{\displaystyle\int_0^2 \frac{x^3\,\dx}{\sqrt{4 - x^2}}}$
$\displaystyle\int x^5\,e^{x}\,\dx\vphantom{\displaystyle\int_0^2 \frac{x^3\,\dx}{\sqrt{4 - x^2}}}$
$\displaystyle\int_0^2 \frac{x^3\,\dx}{\sqrt{4 - x^2}}$
$\displaystyle\int_0^1 x^3\,\sqrt{1+x^2}\,\dx\vphantom{\displaystyle\int_0^2 \frac{x^3\,\dx}{\sqrt{4 - x^2}}}$

5
$\displaystyle\int \sin\,(\ln\,x)~\dx$
$\displaystyle\int \ln\,(1+x^2)\,\dx$
$\displaystyle\int x\,\tan^{-1} x~\dx$
$\displaystyle\int \cot^{-1}\,\sqrt{x}~\dx$
$\displaystyle\int e^{\sqrt{x}}\,\dx$
Evaluate the integral $\int e^x\,\sin\,x\,\dx$ from Example 6.8 by using two rounds of the tabular method and the formula $\int u_1\,\dv_1 = u_1v_1 - u_2v_2 + \int v_2\,\du_2$ from p.162.

Show that for all constants $a$ and $b \ne 0$ :

\[\int e^{ax}\,\cos\,bx~\dx ~=~ \frac{e^{ax}\,(a\,\cos\,bx \;+\; b\,\sin\,bx)}{a^2 + b^2} ~+~ C \enskip\text{and}\enskip \int e^{ax}\,\sin\,bx~\dx ~=~ \frac{e^{ax}\,(a\,\sin\,bx \;-\; b\,\cos\,bx)}{a^2 + b^2} ~+~ C\]
For the Gamma function $\Gamma\,(t)$ show the following:
1. $\Gamma\,(t + 1) ~=~ t\,\Gamma\,(t)$ for all $t > 0$. (Hint: Use integration by parts.)
2. $\Gamma\,(n) ~=~ (n-1)\,!~$ for all positive integers $n$. (Hint: Use part (a) and induction.)
Note that by part (b) the Gamma function can be thought of as an extension of the factorial operation to all positive real numbers. In fact, the Gamma function was created for that purpose.
Use Exercise 28 to prove for all integers $n \ge 1$:

\[ \int_0^{\infty} r^n\,e^{-r}\,\ln\,r~\dr ~=~ (n-1)\,! ~+~ n\,\int_0^{\infty} r^{n-1}\,e^{-r}\,\ln\,r~\dr \]
By the Maxwell speed distribution for gas molecules, the average speed $\avg{\nu}$ of molecules of mass $m$ in a gas at temperature $T$ is

\[ \avg{\nu} ~=~ 4\pi\,\left(\frac{m}{2\pi kT}\right)^{3/2}\; \int_0^{\infty} \nu^3\,e^{-m\nu^2/2kT}\,d\!\nu ~, \]

where $k \approx 1.38056 \times 10^{-23}$ J/K is the Boltzmann constant. Show that

\[ \avg{\nu} ~=~ \sqrt{\frac{8kT}{\pi m}} ~. \]
Some physics texts write integrals in a form like this energy integral from statistical mechanics,

\[ \int_0^{\infty} \ln\,\left(1 - \alpha e^{-x^2}\right)~d(x^3) \]

which uses the differential of a function—in this case $d(x^3)$—instead of a variable (e.g. not just plain $\dx$). This often signals that integration by parts is on the way, with the added benefit of having the $v=\int \dv$ calculation done for you—in the above integral $d(x^3)$ means that $v=x^3$, with $\dv=d(x^3)$ not really being needed for anything else. With that understanding, show that for $0 < \alpha < 1$,

\[ \int_0^{\infty} \ln\,\left(1 - \alpha e^{-x^2}\right)~d(x^3) ~=~ -2\alpha\,\int_0^{\infty} \frac{x^4\,e^{-x^2}\,\dx}{1 - \alpha e^{-x^2}} ~. \]
Find the flaw in the following “proof” that $0 = 1$:

Evaluating the integral $~\displaystyle\int \frac{\dx}{x}~$ using integration by parts with $u = \dfrac{1}{x}$ and $\dv = \dx$, so that $\du = -\dfrac{\dx}{x^2}$ and $v = x$, shows that

\begin{align*} \int u\,\dv ~&=~ uv ~-~ \int v\,\du\\[6pt] \int \frac{\dx}{x} ~&=~ \left(\frac{1}{x}\right) \,\cdot\,x ~-~ \int x \,\cdot\, \left(-\frac{\dx}{x^2}\right)\\[6pt] \int \frac{\dx}{x} ~&=~ 1 ~+~ \int \frac{\dx}{x}\\[6pt] 0 ~&=~ 1 \quad\checkmark \end{align*}

Created by the Swiss mathematician, physicist and astronomer Leonhard Euler (1707-1783). The use of the Greek capital letter $\Gamma$ for this function is due to the French mathematician Adrien-Marie Legendre (1752-1833). ↩