Infinite Integration using the ceiling function $$\int_0^\infty \frac{\sin(x\pi)}{\lceil x \rceil^2 + \lceil x \rceil} dx$$
My teacher recently gave me this and it's stumped me.  
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}{\sin\pars{\pi x} \over
\left\lceil{x}\right\rceil^{2} + \left\lceil{x}\right\rceil}
\,\dd x} =
\sum_{k = 0}^{\infty}\int_{k}^{k + 1}
{\sin\pars{\pi x} \over
\pars{k + 1}^{2} + \pars{k + 1}}\,\dd x
\\[5mm] = &\
{2 \over \pi}\sum_{k = 0}^{\infty}
{\pars{-1}^{k} \over \pars{k + 1}\pars{k + 2}} =
{2 \over \pi}\bracks{%
\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + 1} -
\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + 2}}
\\[5mm] = &\
{2 \over \pi}\bracks{%
-\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k} -
\sum_{k = 2}^{\infty}{\pars{-1}^{k} \over k}}
\\[5mm] = &\
-\,{2 \over \pi}\braces{%
\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k} +
\bracks{1 + \sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k}}} \\[5mm] = &\
-\,{2 \over \pi}\bracks{%
1 + 2\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k}} =
\bbx{4\ln\pars{2} - 2 \over \pi} \approx 0.2459
\end{align}
A: Hint: break the integral up into
$$\int_0^\infty=\int_0^1+\int_1^2+\int_2^3+\cdots.$$
On each of these intervals the denominator takes a constant value, so you can bring it out of the integral sign and evaluate each. After that just sum all the terms you get.
