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Resolución de integrales/ pi*x^2/ 4*pi^2*a*x^2*sin(pi*x^2)

Integral de 4*pi^2*a*x^2*sin(pi*x^2) dx

Límites de integración:

interior superior
v

Gráfico:

interior superior

Definida a trozos:

Solución

Ha introducido [src] 
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∫01x2⋅4π2asin⁡(πx2) dx\int\limits_{0}^{1} x^{2} \cdot 4 \pi^{2} a \sin{\left(\pi x^{2} \right)}\, dx 
Integral((((4*pi^2)*a)*x^2)*sin(pi*x^2), (x, 0, 1))
Solución detallada

  1. Usamos la integración por partes:

    ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

    que u(x)=4π2ax2u{\left(x \right)} = 4 \pi^{2} a x^{2} y que dv⁡(x)=sin⁡(πx2)\operatorname{dv}{\left(x \right)} = \sin{\left(\pi x^{2} \right)}.

    Entonces du⁡(x)=8π2ax\operatorname{du}{\left(x \right)} = 8 \pi^{2} a x.

    Para buscar v(x)v{\left(x \right)}:

      FresnelSRule(a=pi, b=0, c=0, context=sin(pi*x**2), symbol=x)

    Ahora resolvemos podintegral.

  2. La integral del producto de una función por una constante es la constante por la integral de esta función:

    ∫42π2axS(2x) dx=42π2a∫xS(2x) dx\int 4 \sqrt{2} \pi^{2} a x S\left(\sqrt{2} x\right)\, dx = 4 \sqrt{2} \pi^{2} a \int x S\left(\sqrt{2} x\right)\, dx

    1. No puedo encontrar los pasos en la búsqueda de esta integral.

      Pero la integral

      2πx5Γ(34)Γ(54)2F3(34,5432,74,94|−π2x44)16Γ(74)Γ(94)\frac{\sqrt{2} \pi x^{5} \Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{5}{4}\right) {{}_{2}F_{3}\left(\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{3}{2}, \frac{7}{4}, \frac{9}{4} \end{matrix}\middle| {- \frac{\pi^{2} x^{4}}{4}} \right)}}{16 \Gamma\left(\frac{7}{4}\right) \Gamma\left(\frac{9}{4}\right)}

    Por lo tanto, el resultado es: π3ax5Γ(34)Γ(54)2F3(34,5432,74,94|−π2x44)2Γ(74)Γ(94)\frac{\pi^{3} a x^{5} \Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{5}{4}\right) {{}_{2}F_{3}\left(\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{3}{2}, \frac{7}{4}, \frac{9}{4} \end{matrix}\middle| {- \frac{\pi^{2} x^{4}}{4}} \right)}}{2 \Gamma\left(\frac{7}{4}\right) \Gamma\left(\frac{9}{4}\right)}

  3. Ahora simplificar:

    −2π2ax2(4πx32F3(34,5432,74,94|−π2x44)−152S(2x))15- \frac{2 \pi^{2} a x^{2} \left(4 \pi x^{3} {{}_{2}F_{3}\left(\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{3}{2}, \frac{7}{4}, \frac{9}{4} \end{matrix}\middle| {- \frac{\pi^{2} x^{4}}{4}} \right)} - 15 \sqrt{2} S\left(\sqrt{2} x\right)\right)}{15}

  4. Añadimos la constante de integración:

    −2π2ax2(4πx32F3(34,5432,74,94|−π2x44)−152S(2x))15+constant- \frac{2 \pi^{2} a x^{2} \left(4 \pi x^{3} {{}_{2}F_{3}\left(\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{3}{2}, \frac{7}{4}, \frac{9}{4} \end{matrix}\middle| {- \frac{\pi^{2} x^{4}}{4}} \right)} - 15 \sqrt{2} S\left(\sqrt{2} x\right)\right)}{15}+ \mathrm{constant}

Respuesta:

−2π2ax2(4πx32F3(34,5432,74,94|−π2x44)−152S(2x))15+constant- \frac{2 \pi^{2} a x^{2} \left(4 \pi x^{3} {{}_{2}F_{3}\left(\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{3}{2}, \frac{7}{4}, \frac{9}{4} \end{matrix}\middle| {- \frac{\pi^{2} x^{4}}{4}} \right)} - 15 \sqrt{2} S\left(\sqrt{2} x\right)\right)}{15}+ \mathrm{constant}

Respuesta (Indefinida) [src] 
                                                                                                                              
                                                                                                 _  /              |    2  4 \
  /                                                                 3  5                        |_  |   3/4, 5/4   | -pi *x  |
 |                                                              a*pi *x *Gamma(3/4)*Gamma(5/4)* |   |              | --------|
 |     2    2    /    2\                ___   2  2  /    ___\                                  2  3 \3/2, 7/4, 9/4 |    4    /
 | 4*pi *a*x *sin\pi*x / dx = C + 2*a*\/ 2 *pi *x *S\x*\/ 2 / - --------------------------------------------------------------
 |                                                                                 2*Gamma(7/4)*Gamma(9/4)                    
/
∫x2⋅4π2asin⁡(πx2) dx=C−π3ax5Γ(34)Γ(54)2F3(34,5432,74,94|−π2x44)2Γ(74)Γ(94)+22π2ax2S(2x)\int x^{2} \cdot 4 \pi^{2} a \sin{\left(\pi x^{2} \right)}\, dx = C - \frac{\pi^{3} a x^{5} \Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{5}{4}\right) {{}_{2}F_{3}\left(\begin{matrix} \frac{3}{4}, \frac{5}{4} \\ \frac{3}{2}, \frac{7}{4}, \frac{9}{4} \end{matrix}\middle| {- \frac{\pi^{2} x^{4}}{4}} \right)}}{2 \Gamma\left(\frac{7}{4}\right) \Gamma\left(\frac{9}{4}\right)} + 2 \sqrt{2} \pi^{2} a x^{2} S\left(\sqrt{2} x\right)
Simplificar
Respuesta [src] 
        /                      ___  /  ___\           \
      2 |  5*Gamma(5/4)    5*\/ 2 *C\\/ 2 /*Gamma(5/4)|
4*a*pi *|--------------- + ---------------------------|
        \8*pi*Gamma(9/4)         16*pi*Gamma(9/4)     /
4π2a(52C(2)Γ(54)16πΓ(94)+5Γ(54)8πΓ(94))4 \pi^{2} a \left(\frac{5 \sqrt{2} C\left(\sqrt{2}\right) \Gamma\left(\frac{5}{4}\right)}{16 \pi \Gamma\left(\frac{9}{4}\right)} + \frac{5 \Gamma\left(\frac{5}{4}\right)}{8 \pi \Gamma\left(\frac{9}{4}\right)}\right) 
=
= 
        /                      ___  /  ___\           \
      2 |  5*Gamma(5/4)    5*\/ 2 *C\\/ 2 /*Gamma(5/4)|
4*a*pi *|--------------- + ---------------------------|
        \8*pi*Gamma(9/4)         16*pi*Gamma(9/4)     /
4π2a(52C(2)Γ(54)16πΓ(94)+5Γ(54)8πΓ(94))4 \pi^{2} a \left(\frac{5 \sqrt{2} C\left(\sqrt{2}\right) \Gamma\left(\frac{5}{4}\right)}{16 \pi \Gamma\left(\frac{9}{4}\right)} + \frac{5 \Gamma\left(\frac{5}{4}\right)}{8 \pi \Gamma\left(\frac{9}{4}\right)}\right) 
4*a*pi^2*(5*gamma(5/4)/(8*pi*gamma(9/4)) + 5*sqrt(2)*fresnelc(sqrt(2))*gamma(5/4)/(16*pi*gamma(9/4)))

    Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.

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