Calcular esta integral

Question

Tengo problemas para calcular esta integral.

Sea $1 y $a,b\geq 0$. Calcula:

$$ \int_{0}^{x} \frac{a}{(1+bt^{\frac{q}{p-1}})^\frac{q}{q-p}}dt $$

Sacaría el 'a' de la integral. Pero no sabría cómo seguir. ¿Podrías ayudarme?

Answer 1

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Dado $0\le a,0\le b,1, tenemos $0<\frac{q}{p-1}\land0<\frac{q}{q-p}$. Entonces,

$$\begin{align} \int_{0}^{x}\mathrm{d}t\,\frac{a}{\left(1+bt^{\frac{q}{p-1}}\right)^{\frac{q}{q-p}}} &=\int_{0}^{x^{\frac{q}{p-1}}}\mathrm{d}u\,\frac{a\left(\frac{p-1}{q}\right)u^{\frac{p-1}{q}-1}}{\left(1+bu\right)^{\frac{q}{q-p}}};~\small{\left[t^{\frac{q}{p-1}}=u\right]}\\ &=a\mu\int_{0}^{x^{1/\mu}}\mathrm{d}u\,\frac{u^{\mu-1}}{\left(1+bu\right)^{\nu}};~\small{\left[\mu:=\frac{p-1}{q}\land\nu:=\frac{q}{q-p}\right]}\\ &=a\mu\int_{0}^{y}\mathrm{d}u\,\frac{u^{\mu-1}}{\left(1+bu\right)^{\nu}};~\small{\left[y:=x^{1/\mu}>0\right]}\\ &=a\mu\,y^{\mu}\int_{0}^{1}\mathrm{d}t\,\frac{t^{\mu-1}}{\left(1+byt\right)^{\nu}};~\small{\left[u=yt\right]}\\ &=a\mu\,y^{\mu}\operatorname{B}{\left(\mu,1\right)}\,{_2F_1}{\left(\nu,\mu;\mu+1;-by\right)}\\ &=ay^{\mu}\,{_2F_1}{\left(\nu,\mu;\mu+1;-by\right)}\\ &=ax\,{_2F_1}{\left(\frac{q}{q-p},\frac{p-1}{q};\frac{p+q-1}{q};-bx^{\frac{q}{p-1}}\right)},\\ \end{align}$$

donde hemos utilizado la representación integral de Euler para la función hipergeométrica de Gauss que establece

$$\operatorname{B}{\left(\beta,\gamma-\beta\right)}\,{_2F_1}{\left(\alpha,\beta;\gamma;z\right)}=\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta-1}\left(1-t\right)^{\gamma-\beta-1}}{\left(1-zt\right)^{\alpha}};~~~\small{\left(\alpha,\beta,\gamma,z\right)\in\mathbb{R}^{4}\land0<\beta<\gamma\land z<1}.$$

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Answer 2

Por el momento, olvida el $a$ y permita que $$\alpha=\frac{q}{p-1}\qquad \text{y}\qquad \beta=\frac{q}{q-p}$$ y considera el problema de $$I=\int \frac{dt}{(1+b~ t^\alpha)^\beta}$$ $$b~ t^\alpha=u \implies t=b^{-\frac{1}{\alpha } }\,u^{\frac{1}{\alpha }}\implies dt=\frac{b^{-\frac{1}{\alpha } }}{\alpha } \,u^{\frac{1}{\alpha }-1}\,du$$ $$I=\frac{b^{-\frac{1}{\alpha } }}{\alpha } \int u^{\frac{1}{\alpha }-1} (u+1)^{-\beta }\,du=\left(\frac{u}{b}\right)^{\frac{1}{\alpha }} \, _2F_1\left(\frac{1}{\alpha },\beta ;1+\frac{1}{\alpha};-u\right)$$