10
$\begingroup$

I have been working on the indefinite integral of $\cos x/(1+x^2)$.

$$ \int\frac{\cos x}{1+x^2}\;dx\text{ or } \int\frac{\sin x}{1+x^2}\;dx $$

are they unsolvable(impossible to solve) or is there a way to solve them even by approximation?

Thank you very much.

  • 2
    wolfram alpha provides a [solution](http://www.wolframalpha.com/input/?i=Integrate%5BCos%5Bx%5D%2F%281+%2B+x%5E2%29%2C+x%5D) in terms of sine and cosine-integrals.2012-04-09
  • 0
    Why is the word "undefined" in the title?2012-04-09
  • 0
    By maple and matlab also you can get the solution in terms of cosine and sine but i want to know if there is exact solution or the way for it?2012-04-09
  • 0
    Please define what exactly you mean by "unsolvable".2012-04-09
  • 1
    I do not believe "impossible to solve" is a definition in the sense of @Aryabhata. Why do you not accept the solution in terms of sine and cosine-integral as being a solution? What would be a solution for you?2012-04-09
  • 0
    A friend of mine told me that [$\int e^{ix}/(1+x^2)dx=...$](http://tinyurl.com/cppbp4x) Does this help you?2012-05-09
  • 1
    I guess there is no explicit formula for the indefinite integral. I know and estimate $$ \int_{0}^{\pi/2}\frac{\cos x}{1+x^2}\;dx\ge \int_{0}^{\pi/2}\frac{\sin x}{1+x^2}\;dx $$2012-04-09
  • 2
    Defacing your questions is quite frowned upon; please don't do this.2013-03-27

2 Answers 2

11

There is no elementary antiderivative for either of those.

It's actually easier to deal with $e^{ix}/(1+x^2)$. As a corollary of a theorem of Liouville, if $f e^g$ has an elementary antiderivative, where $f$ and $g$ are rational functions and $g$ is not constant, then it has an antiderivative of the form $h e^g$ where $h$ is a rational function. For this to be an antiderivative of $f e^g$, what we need is $h' + h g' = f$.

Now with $f = 1/(1+x^2)$ and $g = ix$, the condition is $h' + i h = 1/(1+x^2)$. The right side has a pole of order $1$ at $x=i$. In order for the left side to have a pole there, $h$ must have a pole there, but wherever $h$ has a pole of order $k$, $h'$ has a pole of order $k+1$, so the left side can never have a pole of order $1$.

5

$\int\dfrac{\sin x}{1+x^2}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!(x^2+1)}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{2(2n+1)!(x^2+1)}d(x^2+1)$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x^2+1-1)^n}{2(2n+1)!(x^2+1)}d(x^2+1)$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^{n-k}(x^2+1)^k}{2(2n+1)!(x^2+1)}d(x^2+1)$

$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(x^2+1)$

$=\int\left(\dfrac{1}{2(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$

$=\int\left(\dfrac{1}{2(x^2+1)}+\sum\limits_{n=1}^\infty\dfrac{1}{2(2n+1)!(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$

$=\int\left(\sum\limits_{n=0}^\infty\dfrac{1}{2(2n+1)!(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$

$=\int\left(\dfrac{\sinh1}{2(x^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(x^2+1)$

$=\dfrac{\sinh1\ln(x^2+1)}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(x^2+1)^k}{2(2n+1)!k!k(n-k)!}+C$