Evaluating $\int_{-3}^{3}\frac{x^8}{1+e^{2x}}dx$

Question

$$\int_{-3}^{3}\frac{x^8}{1+e^{2x}}dx$$

I can't find solution for this task, can someone help me?

Answer 1

Hint. You may write $$ \begin{align} \int_{-3}^{3}\frac{x^8}{1+e^{2x}}dx&=\int_{-3}^0\frac{x^8}{1+e^{2x}}dx+\int_0^{3}\frac{x^8}{1+e^{2x}}dx\\\\ &=\int_0^3\frac{x^8}{1+e^{-2x}}dx+\int_0^{3}\frac{x^8}{1+e^{2x}}dx\\\\ &=\int_0^3\frac{e^{2x}x^8}{1+e^{2x}}dx+\int_0^{3}\frac{x^8}{1+e^{2x}}dx\\\\ &=\int_0^3x^8dx \end{align} $$ and may conclude easily.

Answer 2

The answer from @OlivierOloa presents a really cool trick and I'd like to add some information around it. This might help to apply this technique to similar expressions.

The following holds true. If the integral has the form:

\begin{align*} \int_{-a}^{a}\frac{p(x)}{1+q(x)}\,dx \end{align*}

with

• $p(x)$ is an even function, i.e. $p(x)=p(-x)$

• $q(x)q(-x) = 1$

then

\begin{align*} \int_{-a}^{a}\frac{p(x)}{1+q(x)}\,dx=\frac{1}{2}\int_{-a}^{a}p(x)\,dx \end{align*}

A reasoning is given below. But first let's check our example.

The current integral is symmetric around $x=0$

\begin{align*} \int_{-3}^{3}\frac{x^8}{1+e^{2x}} \end{align*}

with an even function $p(x)=x^8$ and a function $q(x)=e^{-2x}$ with \begin{align*} q(x)q(-x)=e^{-2x}e^{2x}=1 \end{align*}

Now recall that each function $f(x)$ can be uniquely written as sum of an odd function $f_o(x)$ and of an even function $f_e(x)$, since \begin{align*} f(x)&=f_o(x)+f_e(x)\\ f_o(x)&=\frac{1}{2}\left(f(x)-f(-x)\right)\\ f_e(x)&=\frac{1}{2}\left(f(x)+f(-x)\right) \end{align*}

Let's define

\begin{align*} f(x)=\frac{p(x)}{1+q(x)} \end{align*} with $f_e(x)$ it's even and $f_o(x)$ it's odd part.

Since the integral is symmetric around $x=0$ the odd part vanishes and we obtain

\begin{align*} \int_{-a}^{a}f(x)\,dx&=\int_{-a}^{a}f_e(x)\,dx\\ &=\frac{1}{2}\int_{-a}^{a}\left(f(x)+f(-x)\right)\,dx\\ &=\frac{1}{2}\int_{-a}^{a}\left(\frac{p(x)}{1+q(x)}+\frac{p(-x)}{1+q(-x)}\right)\,dx\\ &=\frac{1}{2}\int_{-a}^{a}p(x)\frac{1+q(-x)+1+q(x)}{1+q(x)+q(-x)+q(x)q(-x)}\tag{1}\,dx\\ &=\frac{1}{2}\int_{-a}^{a}p(x)\frac{2+q(x)+q(-x)}{2+q(x)+q(-x)}\tag{2}\,dx\\ &=\frac{1}{2}\int_{-a}^{a}p(x)\,dx \end{align*}

In (1) we use the fact that $p(x)$ is even and in (2) we use $q(x)q(-x)=1$.

$$ $$

In the current situation we obtain \begin{align*} \int_{-3}^{3}\frac{x^8}{1+e^{-2x}}\,dx&=\frac{1}{2}\int_{-3}^{3}\left(\frac{x^8}{1+e^{-2x}}+\frac{x^8}{1+e^{2x}}\right)\,dx\\ &=\frac{1}{2}\int_{-3}^{3}x^8\frac{1+e^{2x}+1+e^{-2x}}{1+e^{-2x}+e^{2x}+e^{-2x}e^{2x}}\,dx\\ &=\frac{1}{2}\int_{-3}^{3}x^8\frac{2+e^{-2x}+e^{2x}}{2+e^{-2x}+e^{2x}}\,dx\\ &=\frac{1}{2}\int_{-3}^{3}x^8\,dx \end{align*}

Note: This technique can be found e.g. in Inside Interesting Integrals written by P.J. Nahin.

He applies this technique to the seemingly complicated integral

\begin{align*} \int_{-1}^{1}\frac{\cos(x)}{1+e^{(1/x)}}\,dx \end{align*}

which becomes easy if you know the trick.

Answer 3

Using $\displaystyle\int_a^bf(y)\ dy=\int_a^bf(a+b-y)\ dy,$

$$I=\int_{-a}^a\frac{x^{2n}}{1+e^{mx}}dx=\int_{-a}^a\frac{(-a+a-x)^{2n}}{1+e^{m(-a+a-x)}}dx=\int_{-a}^a\dfrac{x^{2n}e^{mx}}{1+e^{mx}}dx$$

$$I+I=\int_{-a}^a\frac{x^{2n}}{1+e^{mx}}dx+\int_{-a}^a\dfrac{x^{2n}e^{mx}}{1+e^{mx}}dx=\int_{-a}^ax^{2n}\ dx=?$$

Answer 4

Let $$\displaystyle I =\int_{-3}^{3}\frac{x^8}{1+e^{2x}}dx................(1)\;,$$

Now Let $x=-t\;,$ Then $dx = -dt$ and Changing Limit, We get

$$\displaystyle I = -\int_{3}^{-3}\frac{t^8}{1+e^{-2t}}dt = \int_{-3}^{3}\frac{t^8\cdot e^{2t}}{1+e^{2t}}dt = \int_{-3}^{3}\frac{x^{8}\cdot e^{2x}}{1+e^{2x}}dx$$

So $$\displaystyle I = \int_{-3}^{3}\frac{x^{8}\cdot e^{2x}}{1+e^{2x}}dx.................(2)$$

Above we have used $$\displaystyle \bullet \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$ and the used $$\displaystyle \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$$

So we get $$\displaystyle 2I = \int_{-3}^{3}\frac{x^8\cdot \left(1+e^{2x}\right)}{1+e^{2x}}dx = \int_{-3}^{3}x^8dx = 2\int_{0}^{3}x^8dx = \frac{2\cdot 3^9}{9}$$

So we get $$\displaystyle I = \int_{-3}^{3}\frac{x^{8}}{1+e^{2x}}dx = 3^7$$