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
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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 $$J_1 = x(2\sin x - x\cos x) -2(1-\cos x),\\ J_2 = x^3(4\sin x - x \cos x )-12x(2\sin x - x \cos x)+24(1-\cos x), \dots$$
$$\int_0^x \frac{\sin t}{1 + t^2}\, dt = \int_0^x \sin t\left(1 -t^2 + t^4 + \dots\right) = J_0- J_1 + J_2 + \dots, \text{ for } -1 < x < 1.$$
$\int_0^x\dfrac{\sin(t)}{1+t^2}dt$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n+1}}{(2n+1)!(t^2+1)}dt$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^nt^{2n}}{2(2n+1)!(t^2+1)}d(t^2+1)$
$=\int_0^x\sum\limits_{n=0}^\infty\dfrac{(-1)^n(t^2+1-1)^n}{2(2n+1)!(t^2+1)}d(t^2+1)$
$=\int_0^x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^{n-k}(t^2+1)^k}{2(2n+1)!(t^2+1)}d(t^2+1)$
$=\int_0^x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}d(t^2+1)$
$=\int_0^x\left(\dfrac{1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\int_0^x\left(\dfrac{1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\dfrac{1}{2(2n+1)!(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\int_0^x\left(\sum\limits_{n=0}^\infty\dfrac{1}{2(2n+1)!(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\int_0^x\left(\dfrac{\sinh1}{2(t^2+1)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^{k-1}}{2(2n+1)!k!(n-k)!}\right)d(t^2+1)$
$=\left[\dfrac{\sinh1\ln(t^2+1)}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!(t^2+1)^k}{2(2n+1)!k!k(n-k)!}\right]_0^x$
$=\dfrac{\sinh1\ln(x^2+1)}{2}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^kn!((x^2+1)^k-1)}{2(2n+1)!k!k(n-k)!}$
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 $$\int_0^x \dfrac{\sin(t)}{1+t^2}dt =\sum_{j=0}^{\infty} \frac{(-1)^{j+2}}{(2j+1)!} \frac{x^{4j+2}}{(4j+2)!}$$
