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Usamos la integración por partes:
∫udv=uv−∫vdu
que u(x)=log(x) y que dv(x)=xe−x.
Entonces du(x)=x1.
Para buscar v(x):
EiRule(a=-1, b=0, context=exp(-x)/x, symbol=x)
Ahora resolvemos podintegral.
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No puedo encontrar los pasos en la búsqueda de esta integral.
Pero la integral
⎩⎨⎧−x3F3(1,1,12,2,2−x)+2log(x)2+γlog(x)+iπlog(x)−x3F3(1,1,12,2,2−x)+iπlog(x1)+2log(x)2+γlog(x)+2iπlog(x)−x3F3(1,1,12,2,2−x)+iπG2,22,0(0,01,1x)−iπG2,20,2(1,10,0x)+2log(x)2+γlog(x)+2iπlog(x)for∣x∣<1for∣x∣1<1otherwese
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Ahora simplificar:
⎩⎨⎧x3F3(1,1,12,2,2−x)−2log(x)2+log(x)Ei(−x)−γlog(x)−iπlog(x)x3F3(1,1,12,2,2−x)−iπlog(x1)−2log(x)2+log(x)Ei(−x)−γlog(x)−2iπlog(x)x3F3(1,1,12,2,2−x)+iπ⎩⎨⎧0−log(x1)G2,20,2(1,10,0x)for∣x∣<1for∣x∣1<1otherwese−iπ⎩⎨⎧−log(x)0G2,22,0(0,01,1x)for∣x∣<1for∣x∣1<1otherwese−2log(x)2+log(x)Ei(−x)−γlog(x)−2iπlog(x)for∣x∣<1for∣x∣1<1otherwese
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Añadimos la constante de integración:
⎩⎨⎧x3F3(1,1,12,2,2−x)−2log(x)2+log(x)Ei(−x)−γlog(x)−iπlog(x)x3F3(1,1,12,2,2−x)−iπlog(x1)−2log(x)2+log(x)Ei(−x)−γlog(x)−2iπlog(x)x3F3(1,1,12,2,2−x)+iπ⎩⎨⎧0−log(x1)G2,20,2(1,10,0x)for∣x∣<1for∣x∣1<1otherwese−iπ⎩⎨⎧−log(x)0G2,22,0(0,01,1x)for∣x∣<1for∣x∣1<1otherwese−2log(x)2+log(x)Ei(−x)−γlog(x)−2iπlog(x)for∣x∣<1for∣x∣1<1otherwese+constant