an iterated integral question

Question

This iterated integral is proving harder than I thought. Evaluate by reversing the order of integration: $$ \int_{0}^{1}\left(\int_{y=x}^{\sqrt{x}}\frac{\sin y}{y}dy\right)dx $$

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#00f}{\large% \int_{0}^{1}\bracks{\int_{x}^{\root{x}}{\sin\pars{y} \over y}\,dy}\,\dd x} =\int_{0}^{1}\bracks{\int_{0}^{1}\Theta\pars{y - x}\Theta\pars{\root{x} - y}{\sin\pars{y} \over y}\,dy}\,\dd x \\[3mm]&=\int_{0}^{1}\bracks{\int_{0}^{1}\Theta\pars{y - x}\Theta\pars{x - y^{2}}{\sin\pars{y} \over y}\,dy}\,\dd x= \int_{0}^{1}{\sin\pars{y} \over y}\bracks{\int_{y^{2}}^{y}\dd x}\,\dd y \\[3mm]&= \int_{0}^{1}{\sin\pars{y} \over y}\pars{y - y^{2}}\,\dd y = \int_{0}^{1}\bracks{\sin\pars{y} - y\sin\pars{y}}\,\dd y = \cos\pars{1} + \int_{0}^{1}\bracks{\sin\pars{y} - \cos\pars{y}}\,\dd y \\[3mm]&= \cos\pars{1} + \bracks{-\cos\pars{y} - \sin\pars{y}}_{0}^{1} = \cos\pars{1} + \bracks{-\cos\pars{1} - \sin\pars{1} + 1} =\color{#00f}{\large 1 - \sin\pars{1}} \end{align}