0

I need help with this integral: $$-\int_{-\pi/2}^{\pi/2}{\frac{\sin x}{1+e^x}dx}$$

I know this might feel like me asking you to do my homework, but it isn't. I'm simply stuck in this problem and can't figure how to solve it even after investing a couple of hours. I also don't have anyone to ask this problem, therefore I'm asking this question here.

Tom Himler
  • 2,294
Utkarsh Verma
  • 465
  • 1
    WA says this here $$\frac{1}{2} i \left(, _2F_1\left(-i,1;1-i;-e^{-\pi /2}\right)+, _2F_1\left(-i,1;1-i;-e^{\pi /2}\right)-, _2F_1\left(i,1;1+i;-e^{-\pi /2}\right)-, _2F_1\left(i,1;1+i;-e^{\pi /2}\right)\right)$$ – Dr. Sonnhard Graubner Feb 26 '19 at 13:52
  • I bet this isn't homework, but how did you arrive at this integral? – Zacky Feb 26 '19 at 14:11
  • If the denominator is $$1-e^x$$ use https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 – lab bhattacharjee Feb 26 '19 at 15:03
  • @Zacky It was in a textbook of a friend of mine, meant for a college entrance exam(for IIT-JEE). Though I myself believe this question to be out of context for my level, yet I wanted to ensure if that was the case. – Utkarsh Verma Feb 26 '19 at 16:07
  • @Dr.SonnhardGraubner I had looked up WA already and had no idea what the answer even meant, that's why I ended up asking this on SE. – Utkarsh Verma Feb 26 '19 at 16:08
  • @labbhattacharjee I feel there are chances of a misprint in the textbook, therefore the denominator might have been supposed to be as what you've pointed out. – Utkarsh Verma Feb 26 '19 at 16:10

1 Answers1

1

My initial solution (which I've deleted) was incorrect. Turns out this integral is much more difficult. Here is the work I've done (which is NOT a solution):

\begin{equation} I= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{-\sin(x)}{1 + e^x}\:dx\nonumber \end{equation}

Let us consider a more generalised form: \begin{equation} J(f(x), a) = \int_{-a}^{a} \frac{-f(x)}{1 + e^{x}}\:dx \nonumber \end{equation} Where $f(x)$ is odd, i.e. $f(x) = -f(-x)$. To address this integral we first split into: \begin{equation} J(f(x), a) = \int_{-a}^{0} \frac{-f(x)}{1 + e^{x}}\:dx + \int_{0}^{a} \frac{-f(x)}{1 + e^{x}}\:dx \nonumber \end{equation} Considering the first part, we make the substitution $x \mapsto -x$: \begin{equation} \int_{a}^{0} \frac{-f(-x)}{1 + e^{-x}}-\:dx = \int_{0}^{a} \frac{-f(-x)}{1 + e^{-x}}\:dx \nonumber \end{equation} As $f(x)$ is odd we have $f(-x) = -f(x)$, thus, \begin{equation} \int_{0}^{a} \frac{-f(-x)}{1 + e^{-x}}\:dx = \int_{0}^{a} \frac{f(x)}{1 + e^{-x}}\:dx \nonumber \end{equation} Now, \begin{equation} \frac{1}{1 + e^{-x}} = \frac{e^x}{e^x + 1}\nonumber \end{equation} And so, \begin{equation} \int_{0}^{a} \frac{f(x)}{1 + e^{-x}}\:dx = \int_{0}^{a} \frac{e^xf(x)}{1 + e^{x}}\:dx\nonumber \end{equation} Returning to $J(f(x),a)$ we have \begin{align} J(f(x), a) &= \int_{-a}^{0} \frac{f(x)}{1 + e^{x}}\:dx + \int_{0}^{a} \frac{f(x)}{1 + e^{x}}\:dx \nonumber \\ &= \int_{0}^{a} \frac{e^xf(x)}{1 + e^{x}}\:dx + \int_{0}^{a} \frac{-f(x)}{1 + e^{x}}\:dx \nonumber \\ &= \int_{0}^{a} \frac{e^{x}f(x) - f(x)}{e^x + 1}\:dx = \int_0^a \left(\frac{e^{x} - 1}{e^x + 1}\right)f(x)\:dx \nonumber \end{align} This is probably not of value for this specific integral.

Returning to $I$ we have \begin{equation} I = J\left(\sin(x), \frac{\pi}{2}\right) = \int_0^\frac{\pi}{2} \left(\frac{e^{x} - 1}{e^x + 1}\right)\sin(x)\:dx \nonumber \end{equation}

  • When I was solving this problem, I was stuck at this point, the $frac{e^x-1}{e^x+1}$ form. Can't anything be done after that? – Utkarsh Verma Mar 10 '19 at 03:07