1

Problem

a) $f:[0,1]\rightarrow\Bbb{R}$ is an positive continuous function. Find the value of the intgral: $$\int_0^1\frac{f(x)}{f(x)+f(1-x)}dx$$ b) And compute $$\int_{-\pi}^{\pi}\frac{1}{1+e^{\sin (x)}}\,dx$$

(Hint for task b: Let $f(x)=e^{-\frac{1}{2}\sin (2\pi x-\pi)}$ in task a and use the substisution $u=2\pi*x-\pi$.)

My Work

a) I have realized that $f(1-x)=f(a+b-x)$, but i don`t know how to proceed.

b) I tried to use the hint, I still don't understand how to solve the problem. This is what I end up with: $$\int\frac{e^{-\frac{1}{2}\sin u}}{e^{-\frac{1}{2}\sin u}+e^{\frac{1}{2}\sin u}} $$

Any help is welcome.

qwert
  • 53

4 Answers4

6

HINT:

Show by making the substitution $x\mapsto-x$ that

$$\int_{-\pi}^0 \frac1{1+e^{\sin(x)}}\,dx=\int_0^\pi \frac{e^{\sin(x)}}{1+e^{\sin(x)}}\,dx$$

Can you finish now?

Mark Viola
  • 179,405
5

Replace $x$ with $-z$ and observe that$$\begin{align*}\int\limits_{-\pi}^{\pi}\frac {\mathrm dx}{1+e^{\sin x}} & =\int\limits_{-\pi}^{\pi}\frac {\mathrm dz}{1+e^{-\sin z}}\\ & =\int\limits_{-\pi}^{\pi}\mathrm dz\,\frac {e^{\sin z}}{1+e^{\sin z}}\end{align*}$$Calling the integral $I$ and add the two forms together to get$$\begin{align*}2I & =\int\limits_{-\pi}^{\pi}\mathrm dx\\ & =2\pi\end{align*}$$Thus$$\int\limits_{-\pi}^{\pi}\frac {\mathrm dx}{1+e^{\sin x}}\color{blue}{=\pi}$$

To answer your first question, the integral can be evaluated using a similar trick as above. Make the substitution $x=1-z$ and observe that$$\mathfrak{I}=\int\limits_0^1\mathrm dx\,\frac {f(x)}{f(x)+f(1-x)}=\int\limits_0^1\mathrm dz\,\frac {f(1-z)}{f(1-z)+f(z)}$$Since $z$ is a dummy variable in the second integral, we can combine the two integrals together to get$$\begin{align*}\mathfrak{I} & =\frac 12\int\limits_0^1\mathrm dx\,\frac {f(x)+f(1-x)}{f(1-x)+f(x)}\\ & =\frac 12\end{align*}$$

Frank W
  • 5,897
3

Hint: $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ and $$2I=\int_a^bf(x)+f(a+b-x)dx$$

Soham
  • 9,990
  • 1
    +1 because this covers both parts without having to write the same exposition twice. – J.G. Feb 23 '19 at 16:05
0

This is a "look why i know" answer (instead of a "look what i know" one, which is usually simpler). The hint i always give is to plot the function. A raw plot as the one below is enough:

? plot(x=-Pi,Pi, 1/(1+exp(sin(x))))

0.7309975 |'''''''''''_x""""xx_''''''''''''''''''''''''''''''''''''''''''|
          |         _"         x_                                        |
          |       _"             x                                       |
          |      _                "                                      |
          |     x                  "                                     |
          |    x                    "_                                   |
          |   x                                                          |
          |  x                        "                                  |
          | _                          "                                 |
          |_                            "                                |
          _                              x                               _
          |                               x                              |
          |                                _                            "|
          |                                 _                          " |
          |                                  _                        x  |
          |                                                          x   |
          |                                   "_                    x    |
          |                                     _                  x     |
          |                                      _                "      |
          |                                       x             _"       |
          |                                        "x         _"         |
0.2690025 |.........................................."xx____x"...........|
          -3.141593                                               3.141593
?

We collect from it the information that for $f:[-\pi,\pi]\to\Bbb R$, $f(x)=1/(1+\exp\sin x)$ we have $f(\pm\pi)=1/(1+\exp0)=1/2$ and that "with respect to the $1/2$ shift on the $y$-axis" we (may) get an odd function. So let us set $g(x)=f(x) - 1/2$, and see if it is an odd function, $g(-x)=-g(x)$, which may be a surprise first. So we compute $$ \begin{aligned} g(x) &= \frac 1{1+e^{\sin x}}-\frac 12 \\ &= \frac 12\cdot \frac {1-e^{\sin x}}{1+e^{\sin x}}\ , \\[2mm] &\qquad\text{and} \\[2mm] g(-x) &= \frac 12\cdot \frac {1-e^{-\sin x}}{1+e^{-\sin x}} \\ &= \frac 12 \cdot \frac {e^{\sin x}-1}{e^{\sin x}+1} \cdot \frac {e^{-\sin x}}{e^{-\sin x}} \\ &= \frac 12 \cdot \frac {e^{\sin x}-1}{e^{\sin x}+1} \\ &=-g(x)\ . \end{aligned} $$ Now for the symmetric interval (w.r.t. the origin) $J=[-\pi,\pi]$ we can relate $\int_Jf$ and $\int_J g=0$ easily.

dan_fulea
  • 32,856