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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

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    $\begingroup$ It is not an elementary integral! You can have an answer in terms of the sine and cosine integrals. $\endgroup$ – science Feb 13 '15 at 22:58
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    $\begingroup$ You can also use a power series expansion and integrate term by term since we assume the domain is closed and bounded. $\endgroup$ – Mathemagician1234 Feb 13 '15 at 23:03
  • $\begingroup$ yes, and even for unbounded domains... the improper integrals converge too $\endgroup$ – Asier Calbet Feb 14 '15 at 0:08
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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.

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

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

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  • $\begingroup$ 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? $\endgroup$ – Mathemagician1234 Feb 13 '15 at 23:32
  • $\begingroup$ @Mathemagician1234, why don't you write up the easier method. $\endgroup$ – abel Feb 13 '15 at 23:33
  • $\begingroup$ Don't mind if I do.............. $\endgroup$ – Mathemagician1234 Feb 13 '15 at 23:38
  • $\begingroup$ The series for $\frac1{1+t^2}$ can only be used when $|x|\lt1$. $\endgroup$ – robjohn Feb 14 '15 at 11:12
  • $\begingroup$ @robjohn, yes that is true. so i will include that constraint. robjohn, is the rest of my correct? $\endgroup$ – abel Feb 14 '15 at 11:59
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$\int_0^x\dfrac{\sin(t)}{1+t^2}dt$

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

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

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

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

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

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

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

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

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

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

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

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    $\begingroup$ 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? $\endgroup$ – Mathemagician1234 Feb 15 '15 at 0:04
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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 $$\int_0^x \dfrac{\sin(t)}{1+t^2}dt =\sum_{j=0}^{\infty} \frac{(-1)^{j+2}}{(2j+1)!} \frac{x^{4j+2}}{(4j+2)!}$$

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  • $\begingroup$ Many thanks,Brian! $\endgroup$ – Mathemagician1234 Feb 14 '15 at 0:38
  • $\begingroup$ That $\sum_{j=0}^{x}$ is suspect. What does it mean if $x$ is not a natural number? $\endgroup$ – Jack D'Aurizio Feb 14 '15 at 11:48
  • $\begingroup$ (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! $\endgroup$ – Mathemagician1234 Feb 15 '15 at 0:07
  • $\begingroup$ @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?! $\endgroup$ – Math-fun Feb 15 '15 at 13:19

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