# Showing that $\int_{-n}^{n}{x+\tan{x}\over A +B(x+\tan{x})^{2n}}dx=0$

Where n is an integer, $n\ge1$ and $(A,B)$ just constants

$$I=\int_{-n}^{n}{x+\tan{x}\over A +B(x+\tan{x})^{2n}}dx=0$$

It is obvious that

$$\int_{-n}^{n}x+\tan{x}dx=0$$

Let make a substitution for I $$u=x+\tan{x}\rightarrow du=1+\sec^2{x}dx$$

$$\int_{-n}^{n}{u\over A+Bu^{2n}}{du\over 2+\tan^2{u}}=0$$

I can't find a standard integral of this. I am shrugged at this point on how to continued any further, required some help please

Also note that

$$\int_{-n}^{n}{u\over A+Bu^{2n}}du=0$$

And

$$\int_{-n}^{n}{u\over A+Bu^{2n}}{du\over C+D\tan^{2k}{u}}=0$$

Where A,B,C and D are just constants

$n,k\ge1$ are both integers

• The integrand is an odd function of $x$. The conclusion is immediate. – Paramanand Singh Jun 25 '16 at 10:07
• The trickiest part may be to show that the integrand function is integrable. For instance, what if $A=-B$? – Jack D'Aurizio Jun 25 '16 at 16:35

## 3 Answers

Let:
$$f(x)={x+\tan{x}\over A +B(x+\tan{x})^{2n}}$$ Set $x \mapsto -u$

\begin{align} f(-u)&={-u+\tan{-u}\over A +B(-u+\tan{-u})^{2n}} \\ &={-u-\tan{u}\over A +B(-1)^{2n}(u+\tan{u})^{2n}} \\ &=(-1){u+\tan{u}\over A +B(u+\tan{u})^{2n}} \\ &=-f(u) \end{align} Since x and u are dummy variables, $f(-x)=-f(x)$. Thus $f(x)$ is an odd function.
Your integral can be written as:
\begin{align} I&=\int_{0}^{n}f(x)dx+\int_{0}^{n}f(-x)dx \\ &=\int_{0}^{n}f(x)dx-\int_{0}^{n}f(x)dx \\ &= 0 \end{align} Therefore, we obtain:

$$\int_{-n}^{n}{x+\tan{x}\over A +B(x+\tan{x})^{2n}}dx=0$$

• If the function is integrable on $[-n,n]$. The caveat here is that the integral might not exist for some choices of $A,B$. If for example $A = (1+\tan(1))^2$, $B=-1$ then $f(x) \propto \frac{1}{x-1}$ close to $x=1$ and the integral does not exist. – Winther Jun 27 '16 at 15:53
• @Winther I don't see your point. The value around x=1 and x=-1 will cancel out. The answer will still be zero. – Mc Cheng Jun 28 '16 at 3:52
• No. You do get $0$ if you naively try to compute it, however if $f(x)$ has a singularity at (say) $x=0$ then $\int_{-1}^{1}f(x){\rm d}x$ exists only if both $\int_{-1}^{0}f(x){\rm d}x$ and $\int_{0}^{1}f(x){\rm d}x$ exists. Saying that the integral is zero is similar to saying that the sum $1-1+2-2+3-3+4-4 + \ldots$ is zero while the true answer is that it diverges (well to be fair one can say it's zero in a principal value sense, see e.g. this) – Winther Jun 28 '16 at 4:02

Let us call the integrand as

$$f(x)={x+\tan{x}\over A +B(x+\tan{x})^{2n}}$$

Then it is quite evident that it is an odd function of $x$

$$f(-x)=-f(x)$$

and hence you can easily conclude

$$\int_{-n}^{n}f(x)dx=0$$

HINT:

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

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

Here $a=?,b=?$

and $\tan(-x)=-\tan x$