6
$\begingroup$

Back in my undergrad days I wrote a false proof of the following.

Problem. Prove that $\displaystyle\int_0^{2\pi}\frac{dx}{1+e^{\sin{x}}}=\pi$

Proof. Integrating by parts gives $$ \int_0^{2\pi}\frac{dx}{1+e^{\sin{x}}} = \left.\frac{x}{1+e^{\sin{x}}}\right\vert_0^{2\pi}+\int_0^{2\pi} x\cdot\frac{e^{\sin x}\cos{x}}{(e^{\sin{x}}+1)^2}dx =\pi+\int_0^{2\pi} x\cdot\frac{e^{\sin x}\cos{x}}{(e^{\sin{x}}+1)^2}dx $$ Taking $u=\sin x$ in the last integral gives $$ \int_0^{2\pi} x\cdot\frac{e^{\sin x}\cos{x}}{(e^{\sin{x}}+1)^2}dx =\int_0^0\arcsin u\frac{e^u}{(e^u+1)}du=0 $$ and combining the two equations gives the result. $\Box$

Of course, the problem with this proof is that the equation $x=\arcsin u$ is only valid on $[0,\pi/2]$. However, $\sin{x}$ is invertible on the intervals $[\pi/2,3\pi/2]$ and $[3\pi/2,2\pi]$ so it seems that this problem can be circumvented by splitting the integral up into three integrals and individually applying the $u$-substitution.

Can this proof be salvaged?

Edit: I'm aware that there are other ways to prove this result. I'm mainly concerned with the validity of this proof.

Edit: I've voted up both answers because they give correct proofs. I haven't accepted an answer, however, because neither addresses the issue of breaking up a noninvertible function into seperate integrals where the function is invertible, which was my main reason for posting this question.

$\endgroup$
  • $\begingroup$ Based on your latest edit I have updated my answer. $\endgroup$ – Paramanand Singh Feb 12 '14 at 4:34
6
$\begingroup$

Calculation of original integral $$\int_{0}^{2\pi}\frac{dx}{1 + e^{\sin x}}$$ can be done directly using the hint from lab bhattacharjee. I believe you would want to have a proof that the complicated integral $$I = \int_{0}^{2\pi}x\frac{e^{\sin x}\cos x}{(1 + e^{\sin x})^{2}}\,dx = 0$$

To that end let's apply the hint (again) from lab bhattacharjee to get $$I = \int_{0}^{2\pi}(2\pi - x)\frac{e^{\sin x}\cos x}{(1 + e^{\sin x})^{2}}\,dx$$ so that by adding these two equivalent forms of $I$ we get $$I = \pi\int_{0}^{2\pi}\frac{e^{\sin x}\cos x}{(1 + e^{\sin x})^{2}}\,dx = \pi\left(\int_{0}^{\pi}\frac{e^{\sin x}\cos x}{(1 + e^{\sin x})^{2}}\,dx + \int_{\pi}^{2\pi}\frac{e^{\sin x}\cos x}{(1 + e^{\sin x})^{2}}\,dx\right)$$ or $I = \pi(I_{1} + I_{2})$.

Now we can put $t = x - \pi$ in second integral $I_{2}$ to get $$I_{2} = -\int_{0}^{\pi}\frac{e^{\sin t}\cos t}{(1 + e^{\sin t})^{2}}\,dt = -I_{1}$$ It now follows that $I = \pi(I_{1} + I_{2}) = 0$.

Update: After the edit by OP, it is clear that what is needed here is to apply the substitution $u = \sin x$ and then show that the integral $I$ is $0$. As he has mentioned in his question this would need a split into three integrals over the range $[0, \pi/2], [\pi/2, 3\pi/2]$ and $[3\pi/2, 2\pi]$. After the substitution the intervals will change to $[0, 1], [-1, 1]$ and $[-1, 0]$. On doing this substitution it will be found that the integral is equal to $$I = I_{1} + I_{2} + I_{3} = \int_{0}^{1}\frac{e^{u}\arcsin u}{(e^{u} + 1)^{2}}\,du - \int_{-1}^{1}\frac{e^{u}(\pi + \arcsin u)}{(e^{u} + 1)^{2}}\,du + \int_{-1}^{0}\frac{e^{u}(2\pi + \arcsin u)}{(e^{u} + 1)^{2}}\,du$$

Note that in the interval $[0, \pi/2]$ the function $u = \sin x$ increases and hence the mapping $\sin x = u$ can be inverted by $x = \arcsin u$ and it maps $[0, \pi/2]$ into $[0, 1]$. In the interval $[\pi/2, 3\pi/2]$ the function $u = -\sin x$ increases and the inverse happens using $x = (\pi + \arcsin u)$ and since $-\cos x \, dx = du$ we get a $-$ sign in the integral. Again in the interval $[3\pi/2, 2\pi]$ the function $\sin x$ increases and the correct inverse is $x = 2\pi + \arcsin u$. Since the function $e^{u}/(e^{u} + 1)^{2}$ is even we have $$\begin{aligned}I &= \int_{-1}^{0}\frac{e^{u}\arcsin u}{(e^{u} + 1)^{2}}\,du + \int_{0}^{1}\frac{e^{u}\arcsin u}{(e^{u} + 1)^{2}}\,du - \int_{-1}^{1}\frac{e^{u}\arcsin u}{(e^{u} + 1)^{2}}\,du\\ &\,\,\,+\,\,2\pi\int_{-1}^{0}\frac{e^{u}}{(e^{u} + 1)^{2}}\,du - \pi\int_{-1}^{1}\frac{e^{u}}{(e^{u} + 1)^{2}}\,du = 0\end{aligned}$$

$\endgroup$
  • $\begingroup$ Nice solution!! +1 for you my friend! $\endgroup$ – Mark Viola Oct 21 '15 at 4:24
  • $\begingroup$ Comment Part 1: Referring to your update, the interval $[\frac{\pi}{2}$, $\frac{3\pi}{2}]$ has been analysed wrongly: $I_{2}$ should be $$\int_{1}^{-1}\frac{e^{u}(\pi - \arcsin u)}{(e^{u} + 1)^{2}}\,du$$ instead. This is because as $x$ is increasing from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and $u = \sin x$ correspondingly decreasing from $1$ to $-1$, $x$ ought to be given by $x = (\pi - \arcsin u)$. $\endgroup$ – Ryan Nov 12 '17 at 12:58
  • $\begingroup$ Comment Part 2: Therefore, the final equation in your answer should instead be $$\begin{aligned}I &= \int_{-1}^{0}\frac{e^{u}\arcsin u}{(e^{u} + 1)^{2}}\,du + \int_{0}^{1}\frac{e^{u}\arcsin u}{(e^{u} + 1)^{2}}\,du + \int_{-1}^{1}\frac{e^{u}\arcsin u}{(e^{u} + 1)^{2}}\,du\\ &\,\,\,+\,\,2\pi\int_{-1}^{0}\frac{e^{u}}{(e^{u} + 1)^{2}}\,du - \pi\int_{-1}^{1}\frac{e^{u}}{(e^{u} + 1)^{2}}\,du\\ &\,\,\, = 2 \int_{-1}^{1}\frac{e^{u}\arcsin u}{(e^{u} + 1)^{2}}\,du\\ &\,\,\, = 0\end{aligned}$$ since the last integrand is an odd function. $\endgroup$ – Ryan Nov 12 '17 at 13:04
  • $\begingroup$ @Ryan: your substitution is also correct as well as mine. In $I_{2}$ I have used $u=-\sin x$ whereas you have used $u=\sin x$. I don't see any problem with any of the approaches. $\endgroup$ – Paramanand Singh Nov 12 '17 at 13:46
2
$\begingroup$

Not sure how you have reached at $$\left.\frac{x}{1+e^{\sin{x}}}\right\vert_0^{2\pi}=\pi$$

Here is another way:

$$\text{Use }I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$

$$2I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b[f(x)+f(a+b-x)]dx$$

utilizing $$\displaystyle\sin(2\pi+0-x)=-\sin x$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.