# Evaluating $\int_0^x\frac{\sin(t)}{1+t^2}\mathrm dt$

I cannot seem to figure out how to solve the following problem: $$\int_0^x\frac{\sin(t)}{1+t^2}\mathrm dt$$

I have tried by using integration by parts.

I set $u = \sin(t)$, $v = \tan^{-1}(x)$ and $dv = \dfrac{1}{1+t^2}dt$. Thus,

$$\int_0^x \dfrac{\sin(t)}{1+t^2}dt = \left[\sin(t)\cdot \tan^{-1}(t)\right]_0^x - \int_0^x \cos(t)\cdot \tan^{-1}(t)dt$$

Is this correct or not? I would like to get some hints, I am currently stuck.

Best regards

• It is not an elementary integral! You can have an answer in terms of the sine and cosine integrals. Commented Feb 13, 2015 at 22:58
• You can also use a power series expansion and integrate term by term since we assume the domain is closed and bounded. Commented Feb 13, 2015 at 23:03
• yes, and even for unbounded domains... the improper integrals converge too Commented Feb 14, 2015 at 0:08

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{0}^{x}\frac{\sin\pars{t}}{1 + t^{2}}\,\dd t} =\Im\int_{0}^{x}\frac{\sin\pars{t}}{t - \ic}\,\dd t =\Im\int_{-\ic}^{x - \ic}\frac{\sin\pars{t + \ic}}{t}\,\dd t \\[5mm]&=\Im\int_{-\ic}^{x - \ic}\frac{\sin\pars{t}\cosh\pars{1} + \cos\pars{t}\sinh\pars{1}\ic}{t}\,\dd t \\[1cm]&=\cosh\pars{1}\bracks{\Im\int_{0}^{x - \ic}\frac{\sin\pars{t}}{t}\,\dd t -\Im\int_{0}^{-\ic}\frac{\sin\pars{t}}{t}\,\dd t} \\[5mm]&+\sinh\pars{1}\bracks{% -\Re\int_{0}^{x - \ic}\frac{1 - \cos\pars{t}}{t}\,\dd t +\Re\int_{0}^{-\ic}\frac{1 - \cos\pars{t}}{t}\,\dd t +\Re\int_{-\ic}^{x - \ic}\frac{\dd t}{t}} \\[1cm]&=\cosh\pars{1}\bracks{% \Im\,{\rm Si}\pars{x - \ic} - \Im\,{\rm Si}\pars{-\ic}} \\[5mm]&+\sinh\pars{1}\bracks{% -\Re\,{\rm Cin}\pars{x - \ic} + \Re\,{\rm Cin}\pars{-\ic} +\Re\ln\pars{1 + x\ic}} \\[1cm]&=\color{#66f}{\large\cosh\pars{1}\bracks{% \Im\,{\rm Si}\pars{x - \ic} - \Im\,{\rm Si}\pars{-\ic}}} \\[5mm]&\color{#66f}{\large+\sinh\pars{1}\bracks{% -\Re\,{\rm Cin}\pars{x - \ic} + \Re\,{\rm Cin}\pars{-\ic}} + \half\,\sinh\pars{1}\ln\pars{1 + x^{2}}} \end{align}

$\ds{\,{\rm Si}}$ and $\ds{\,{\rm Cin}}$ are the Sine and Cosine Integral Functions, respectively.

i will use series expansion and find a recurrence relation. first define
$$J_k = \int_0^x t^{2k}\sin t \,dt, k = 0, 1, \dots$$

we have $J_0 = \int_0^x \sin \, dt = 1 - \cos x.$ now,

\begin{align} J_k &= \int_0^x t^{2k} \sin t \, dt = -t^{2k}\cos t|_0^x + \int_0^x 2k t^{2k-1} \cos t\, dt\\ &= 2kt^{2k-1}\sin t-t^{2k}\cos t|_0^x - 2k(2k-1)\int_0^x t^{2k-2} \sin t\, dt\\ &= 2kx^{2k-1}\sin x - x^{2k}\cos x -2k(2k-1)J_{k-1} = x^{2k-1}(2k\sin x - x\cos x) -2k(2k-1)J_{k-1} \end{align}

the recurrence relation is $$J_k = x^{2k-1}(2k\sin x - x\cos x) -2k(2k-1)J_{k-1}\, , \, J_0 = 1 - \cos x\tag 1$$

we can compute $$J_1 = x(2\sin x - x\cos x) -2(1-\cos x),\\ J_2 = x^3(4\sin x - x \cos x )-12x(2\sin x - x \cos x)+24(1-\cos x), \dots$$

$$\int_0^x \frac{\sin t}{1 + t^2}\, dt = \int_0^x \sin t\left(1 -t^2 + t^4 + \dots\right) = J_0- J_1 + J_2 + \dots, \text{ for } -1 < x < 1.$$

• Ok,that seems overly complicated to me. Wouldn't it be simpler just to use a straight up power series expansion and integrate term by term? Commented Feb 13, 2015 at 23:32
• @Mathemagician1234, why don't you write up the easier method.
– abel
Commented Feb 13, 2015 at 23:33
• Don't mind if I do.............. Commented Feb 13, 2015 at 23:38
• The series for $\frac1{1+t^2}$ can only be used when $|x|\lt1$.
– robjohn
Commented Feb 14, 2015 at 11:12
• @robjohn, yes that is true. so i will include that constraint. robjohn, is the rest of my correct?
– abel
Commented Feb 14, 2015 at 11:59

$\int_0^x\dfrac{\sin(t)}{1+t^2}dt$

$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n+1}}{(2n+1)!(t^2+1)}dt$

$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n}}{2(2n+1)!(t^2+1)}d(t^2+1)$

$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^n(t^2+1-1)^n}{2(2n+1)!(t^2+1)}d(t^2+1)$

$=\int_0^x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^{n-k}(t^2+1)^k}{2(2n+1)!(t^2+1)}d(t^2+1)$

$=\int_0^x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(t^2+1)$

$=\int_0^x\left(\dfrac{1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$

$=\int_0^x\left(\dfrac{1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\dfrac{1}{2(2n+1)!(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$

$=\int_0^x\left(\sum\limits_{n=0}^\infty\dfrac{1}{2(2n+1)!(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$

$=\int_0^x\left(\dfrac{\sinh1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$

$=\left[\dfrac{\sinh1\ln(t^2+1)}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^k}{2(2n+1)!k!k(n-k)!}\right]_0^x$

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

• Uh-ok,that looks really arduous. I need to spend a few minutes carefully going through the computation. Is mine below correct?I'm pretty sure it is,although of course it's nowhere near as detailed as yours. Is the sum below equivalent? Commented Feb 15, 2015 at 0:04

The relevant power series expansions in the radii of convergence (0,x) are $$\sin (x) = \sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!} x^{2j+1}$$ $$\frac{1}{1+{t^2}} = 1+x^2+x^4+\ldots= \sum_{j=0}^{\infty} {x^{2j}}$$

So now we multiply the power series and integrate term by term: $$\frac{\sin(x)}{1+{t^2}}$$ = $\sum_{j=0}^{\infty} {x^{2j}}\sum_{j=0}^{\infty} \frac{(-1)^j}{(2j+1)!} x^{2j+1}$ = $\sum_{j=0}^{\infty} \frac{(-1)^{j+2}}{(2j+1)!} x^{4j+1}$

We obtained the last part by adding together like coeffecients as follows: 1 = $-1^{2}$ and $\sum_{j=0}^{\infty} {x^{2j}}\cdot x^{2j+1}$= $\sum_{j=0}^{\infty} {x^{4j+1}}$. So now we simply integrate term by term,which yields $$\int_0^x \dfrac{\sin(t)}{1+t^2}dt =\sum_{j=0}^{\infty} \frac{(-1)^{j+2}}{(2j+1)!} \frac{x^{4j+2}}{(4j+2)!}$$

• Many thanks,Brian! Commented Feb 14, 2015 at 0:38
• That $\sum_{j=0}^{x}$ is suspect. What does it mean if $x$ is not a natural number? Commented Feb 14, 2015 at 11:48
• (0,x) is the radii of convergence, obviously-we take each antiderivative term along this radii, but it's overall a convergent countably infinite sum. I rectified it. In light of Harry Peter's intimidating computation above,thought-I'm wondering now if it's correct. It sure LOOKS correct to me! Commented Feb 15, 2015 at 0:07
• @Mathemagician1234 $\sum_{j=0}^{\infty} {x^{2j}}=\frac{1}{1-x^2}$; Set $x=1$ then $\int_0^1 \dfrac{\sin(t)}{1+t^2}dt=0.499769...$ but $\sum_{j=0}^{\infty} \frac{(-1)^{j+2}}{(2j+1)!} \frac{1^{4j+2}}{(4j+2)!}=0.321794$... . Maybe something is wrong?! Commented Feb 15, 2015 at 13:19