Good night i have problem solving this integral.

$$\int_1^2 \int_{\sqrt{x}}^x \sin\left(\frac{\pi x}{2y}\right) \,dy \, dx$$

I make the area of integration, but i cannot solve the integrat, i don't know how! please help me.

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How i can make this??

I think in this form:

$$ \sin\left(\frac{\pi x}{2y}\right)=\sqrt{\frac{1-\cos(\pi x)}{2y}}$$ but i don't know...

  • $\begingroup$ That last re-write: Are you sure it is right? Usually when sine is rewritten with a squareroot and a cosine the cosine inside is squared, but one can get rid of that using a double angle formula (which it looks like you didn't do here). Also suggest you make the capital pi's into lower case, and replace your "sen" by sin in the initial integral. $\endgroup$ – coffeemath May 24 '16 at 23:01
  • $\begingroup$ oh sorry man, @coffeemath i'm spanish and we say "sen", but i go to change to "sin" thanks! $\endgroup$ – Bvss12 May 24 '16 at 23:03
  • $\begingroup$ Bvss12 What about my question regarding your trig replacement formula at the end? [Sorry I didn't know Spanish for sine is "sen"--] Also, it may help to reverse the integration order, since integrating sine of $(x/k)$ where $k$ is a fixed veriable is easy. $\endgroup$ – coffeemath May 24 '16 at 23:05
  • $\begingroup$ you're in the right! oh! you say change the integration limits? @coffeemath $\endgroup$ – Bvss12 May 24 '16 at 23:12
  • $\begingroup$ It tuned out that, even though changing the order made the first integration simpler, the integral had to be4 broken up at $\sqrt{2}$ and the upper part involved an integral of $K\cdot y \cos(k/y),$ which isn't simple. I think you should use Felix Marin's answer, which involves the Ci(x) function. $\endgroup$ – coffeemath May 25 '16 at 10:30

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{y = {\pi x \over 2t}}$: \begin{align} &\color{#f00}{% \int_{1}^{2}\int_{\root{x}}^{x}\sin\pars{{\pi x \over 2y}}\,\dd y\,\dd x} = -\,{\pi \over 2}\int_{1}^{2}x \int_{\pi\root{x}/2}^{\pi/2}{\sin\pars{t} \over t^{2}}\,\dd t\,\dd x \\[3mm] = &\ -\,{\pi \over 2}\int_{1}^{2}x\braces{% \left.-\,{\sin\pars{t} \over t}\right\vert_{\pi\root{x}/2}^{\pi/2} + \int_{\pi\root{x}/2}^{\pi/2}{\cos\pars{t} \over t}\,\dd t}\,\dd x \\[3mm] = &\ -\,{\pi \over 2}\int_{1}^{2}x\bracks{% -\,{1 \over \pi/2} + {\sin\pars{\pi\root{x}/2} \over \pi\root{x}/2} + \mathrm{Ci}\pars{{\pi \over 2}} - \mathrm{Ci}\pars{{\pi\root{x} \over 2}}} \,\dd x \end{align}

where $\mathrm{Ci}$ is the Cosine Integral function: $\ds{\mathrm{Ci}\pars{x} \equiv -\int_{x}^{\infty}{\cos\pars{t} \over t} \,\dd t}$. Then,

\begin{align} &\color{#f00}{% \int_{1}^{2}\int_{\root{x}}^{x}\sin\pars{{\pi x \over 2y}}\,\dd y\,\dd x} \\[3mm] = &\ {3 \over 2} - {3\pi \over 4}\,\mathrm{Ci}\pars{{\pi \over 2}} - \int_{1}^{2}\root{x}\sin\pars{{\pi\root{x} \over 2}}\,\dd x + {\pi \over 2}\int_{1}^{2}x\,\mathrm{Ci}\pars{{\pi\root{x} \over 2}}\,\dd x \\[3mm] = &\ {3 \over 2} - {3\pi \over 4}\,\mathrm{Ci}\pars{{\pi \over 2}} - {16 \over \pi^{3}}\int_{\pi/2}^{\root{2}\pi/2}t^{2}\sin\pars{t}\,\dd t + {16 \over \pi^{3}}\int_{\pi/2}^{\root{2}\pi/2}t^{3}\,\mathrm{Ci}\pars{t}\,\dd t \end{align}

The last integral can be integrated by parts such that we can use $\ds{\mathrm{Ci}'\pars{x} = {\cos\pars{x} \over x}}$.


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