Evaluate $\int_0^e{W(x)}\,\mathrm{d}x$

Question

The function $W(x)$ satisfies $W(x)e^{W(x)}=x$ for all $x$. Evaluate $$\int_0^e{W(x)}\,\mathrm{d}x$$

I tried integrating $xe^{-W(x)}$ but can't see how to do it.

user85798 · Accepted Answer · 2013-11-04T07:23:51.377

8

Integrate by parts with $u=W(x)$ and $\mathrm{d}v=\mathrm{d}x$

$$\int{W(x)}\,\mathrm{d}x=xW(x) - \int xW'(x)\,\mathrm{d}x=xW(x)-\int W(x)W'(x)e^{W(x)}\,\mathrm{d}x$$

Integrate by parts again with $u = W(x)$ and $\mathrm{d}v=W'(x)e^{W(x)}\,\mathrm{d}x$

$$\int W(x)W'(x)e^{W(x)}\,\mathrm{d}x=W(x)e^{W(x)} - \int W'(x)e^{W(x)}\,\mathrm{d}x=W(x)e^{W(x)}-e^{W(x)}$$

So $$\int{W(x)}\,\mathrm{d}x=xW(x) -W(x)e^{W(x)}+e^{W(x)}=x\left(W(x)-1+\frac{1}{W(x)}\right)$$

Note that $W(e) = 1$ and $W(0)=0$ and $\displaystyle{\lim_{x\to 0}\frac{x}{W(x)}=1}$

$$\left|x\left(W(x)-1+\frac{1}{W(x)}\right)\right|_0^e=e\left(W(e)-1+\frac{1}{W(e)}\right) - \left(\lim_{x\to 0}\frac{x}{W(x)}\right) = e-1$$

edited Nov 04 '13 at 07:23

answered Nov 04 '13 at 07:18

user85798

1

Please be more careful in your final step. If you find yourself handling $\frac{0}{0}$, revert to taking the limit as $x\rightarrow 0$ instead. – Bennett Gardiner Nov 04 '13 at 07:23
Yeah good point, I changed it – user85798 Nov 04 '13 at 07:24
Cool, but more specifically, you'll need to use L'Hopital's rule to evaluate this limit. So $$ \lim_{x\rightarrow0} \frac{x}{W(x)} = \lim_{x\rightarrow0} \frac{1}{W'(x)}. $$ Now I believe $$ W'(x) = \frac{\mathrm{e}^{-W(x)}}{W(x)+1},$$ which at $x=0$ evaluates to 1, so you obtain your result. – Bennett Gardiner Nov 04 '13 at 07:27
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@BennettGardiner Why L'Hopital? Note that $x/W(x)=e^{W(x)}$ and $W(x)\to0$ when $x\to0$ hence $x/W(x)\to e^0=1$. – Did Nov 04 '13 at 07:34
Yes, I just rearranged to get $e^0 = 1$ – user85798 Nov 04 '13 at 07:49
Ah, of course! Never mind me then. – Bennett Gardiner Nov 04 '13 at 08:40

score 1 · Answer 2 · answered Feb 20 '16 at 13:07

Let $u = \mbox{W}(x)$. Then the defining equation $\mbox{W}(x) e^{\mbox{W}(x)} = x$ becomes $u e^u = x$ so that $dx = (u + 1)e^u \, du$.

For the limits of integration, when $x = 0, u= \mbox{W}(0)$. From the defining equation we have $\mbox{W}(0) e^{\mbox{W}(0)} = 0 \Rightarrow x = 0$. Also, when $x = e, u = \mbox{W}(e)$ and from the defining equation we have $\mbox{W}(e) e^{\mbox{W}(e)} = e \Rightarrow \mbox{W}(e) = 1$. Thus \begin{align*} \int^e_1 \mbox{W}(x) \, dx &= \int^1_0 u (u + 1) e^u \, du\\ &= \big{[} e^u (u^2 - u + 1) \big{]}^1_0 \quad \mbox{(by parts)}\\ &= e - 1 \end{align*}
Comment: The function $\mbox{W}(x)$ is nothing more than the principal branch of the Lambert W function, $\mbox{W}_0 (x)$.

score 0 · Answer 3 · answered Nov 04 '13 at 07:14

The relation $W(x) e^{W(x)}=x$ means you want to manipulate the integrand in a way which makes an "x" term pop up, which you can then replace with the more complicated expression. You then want this to open up a substitution which gives you a standard integral to evaluate.

An integration by parts is in order.

Evaluate $\int_0^e{W(x)}\,\mathrm{d}x$

3 Answers3