Indefinite Integral for $\cos x/(1+x^2)$

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.

• wolfram alpha provides a solution in terms of sine and cosine-integrals. Apr 9, 2012 at 16:20
• Why is the word "undefined" in the title? Apr 9, 2012 at 16:29
• I guess there is no explicit formula for the indefinite integral. I know and estimate $$\int_{0}^{\pi/2}\frac{\cos x}{1+x^2}\;dx\ge \int_{0}^{\pi/2}\frac{\sin x}{1+x^2}\;dx$$ Apr 9, 2012 at 16:59
• 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? Apr 9, 2012 at 17:27
• Defacing your questions is quite frowned upon; please don't do this. Mar 27, 2013 at 10:33

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 $1$.

• Why you consider only $e^{ix}$? cos(z) = $(e^{iz} + e^{-iz})/2$ Jan 9, 2020 at 16:11
• In this case, since $1/(1+x^2)$ is real, $\int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}dx$ would just be the real part of $\int_{-\infty}^{\infty}\frac{e^{ix}}{1+x^2}dx$ Sep 30, 2022 at 17:19

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

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

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

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

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

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

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

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

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

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

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

If you take $$f(z) = \frac{e^{iz}}{1+z^2}$$ and apply the residue theorem over the path $$\Gamma = \gamma_1 + \gamma_2$$: $\Gamma$" />

You have as follow:

$$\frac{1}{2\pi i}\int_{\Gamma} f(z)dz = \text{Res}(f,i)$$

'cause, of the two poles of $$f(z)$$, only $$i$$ is inside $$\Gamma$$. So, $$\int_{\Gamma} f(z)dz = \int_{\gamma_1} f(z)dz + \int_{\gamma_2} f(z)dz$$. Let's do it one by one:

$$\int_{\gamma_1} f(z)dz = \int_{\gamma_1} \frac{e^{iz}}{1+z^2}dz = \int_{-R}^{R} \frac{e^{it}}{1+t^2}dt$$

because we have parametrized $$\gamma_1$$ as $$\gamma_1(t) = t$$, for $$t\in[-R,R]$$ and $$dz = dt$$.

$$\int_{-R}^{R} \frac{e^{it}}{1+t^2}dt = \int_{-R}^{R} \frac{\cos(t) + i\sin(t)}{1+t^2}dt = \int_{-R}^{R} \frac{\cos(t)}{1+t^2}dt + i\int_{-R}^{R} \frac{\sin(t)}{1+t^2}dt$$ But the second one of this two integrals is zero, cause $$\frac{\sin(t)}{1+t^2}$$ is an odd function. Allright, let's take care of the integral

$$\int_{\gamma_2} f(z)dz = \int_{0}^{\pi} \frac{e^{iRe^{it}}}{1 + R^2e^{i2t}}iRe^{it} dt$$

'cause, now, $$\gamma_2 = e^{it}$$ for $$t \in [0,\pi]$$, so $$z=Re^{it}$$ and $$dz = iRe^{it}dt$$. Then, we have:

$$\int_{0}^{\pi} \frac{e^{iRe^{it}}}{1 + R^2e^{i2t}}iRe^{it} dt = \int_{0}^{\pi} \frac{e^{iR(\cos(t)+i\sin(t))}}{1 + R^2e^{i2t}}iRe^{it}dt =\\ \int_{0}^{\pi} \frac{e^{iR\cos(t)} e^{-R\sin(t)}}{1 + R^2e^{i2t}}iRe^{it}dt \leq \int_{0}^{\pi} \left|\frac{e^{iR\cos(t)} e^{-R\sin(t)}}{1 + R^2e^{i2t}}iRe^{it}dt\right| = \int_{0}^{\pi} \frac{R}{1 + R^2e^{i2t}}e^{-R\sin(t)}dt$$ and, $$\lim_{R\rightarrow +\infty} \int_{0}^{\pi} \frac{R}{1 + R^2e^{i2t}}e^{-R\sin(t)}dt = 0$$

So, summing up, we have

$$\int_{\Gamma} f(z)dz = \int_{-R}^{R} \frac{\cos(t)}{1+t^2}dt$$

and, trivially,

$$\lim_{R\rightarrow +\infty} \int_{-R}^{R} \frac{\cos(t)}{1+t^2}dt = \int_{-\infty}^{\infty} \frac{\cos(t)}{1+t^2}dt$$

just, what we want to know. The computing of Res($$f, i$$) it's simple 'cause $$i$$ is a pole of degree 1, so you just have to derive the denominator and evaluate all in $$z=i$$, And finally, for the residue theorem, we have, for $$R\rightarrow +\infty$$:

$$\frac{1}{2\pi i} \int_{\Gamma} f(z)dz = \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{\cos(t)}{1+t^2}dt = Res(f, i) = \frac{e^{it}}{(1+z^2)'}\Bigg\rvert_{z=i} = \frac{e^{it}}{2z}\Bigg\rvert_{z=i} =\frac{e^{-1}}{2i}\\ \longrightarrow \int_{-\infty}^{\infty} \frac{\cos(t)}{1+t^2}dt = \frac{\pi}{e}$$

• youtu.be/S9LttmTD_14, have a look at this video. He provides a simple solution to it using feynman's trick. It's really simple when compared to other solutions to it. (Complex Analysis also makes toughest integrals simple but it's actually complex). Sep 19, 2021 at 12:45