Resolver esta pregunta Me salió esto (pero mathematica me da una respuesta diferente, también cuando uso un valor para $u$ ) ¿alguien puede detectar mi error?
Mi trabajo:
$$\text{I}(u)=\int_u^\infty\frac{\sqrt{x}}{x^2-2x+2}\space\text{d}x=\lim_{n\to\infty}\int_u^n\frac{\sqrt{x}}{x^2-2x+2}\space\text{d}x=$$
Sustituir $s=\sqrt{x}$ y $\text{d}s=\frac{1}{2\sqrt{x}}\space\text{d}x$ .
Esto da un nuevo límite inferior $s=\sqrt{u}$ y el límite superior $s=\sqrt{n}$ :
$$2\lim_{n\to\infty}\int_{\sqrt{u}}^{\sqrt{n}}\frac{s^2}{s^4-2s^2+2}\space\text{d}s=2\lim_{n\to\infty}\int_{\sqrt{u}}^{\sqrt{n}}\frac{s^2}{\left(s^2-(1+i)\right)\left(s^2-(1-i)\right)}\space\text{d}s=$$
Usar fracciones parciales, puse:
- $$\text{z}=\frac{1+i}{2}\to\overline{\text{z}}=\frac{1-i}{2}$$
-
$$\text{q}=1+i\to\overline{\text{q}}=1-i$$
$$2\lim_{n\to\infty}\int_{\sqrt{u}}^{\sqrt{n}}\left[\frac{\text{z}}{s^2-\overline{\text{q}}}+\frac{\overline{\text{z}}}{s^2-\text{q}}\right]\space\text{d}s=$$ $$2\lim_{n\to\infty}\left[\text{z}\cdot\int_{\sqrt{u}}^{\sqrt{n}}\frac{1}{s^2-\overline{\text{q}}}\space\text{d}s+\overline{\text{z}}\cdot\int_{\sqrt{u}}^{\sqrt{n}}\frac{1}{s^2-\text{q}}\space\text{d}s\right]=$$
Sustituir, para la integral de la izquierda $t=\frac{s}{\sqrt{\overline{\text{q}}}}$ y $\text{d}t=\frac{1}{\sqrt{\overline{\text{q}}}}\space\text{d}s$ .
Esto da un nuevo límite inferior $t=\frac{\sqrt{u}}{\sqrt{\overline{\text{q}}}}$ y el límite superior $t=\frac{\sqrt{n}}{\sqrt{\overline{\text{q}}}}$ :
Sustituir, para la integral de la derecha $v=\frac{s}{\sqrt{\text{q}}}$ y $\text{d}v=\frac{1}{\sqrt{\text{q}}}\space\text{d}s$ .
Esto da un nuevo límite inferior $v=\frac{\sqrt{u}}{\sqrt{\text{q}}}$ y el límite superior $v=\frac{\sqrt{n}}{\sqrt{\text{q}}}$ :
$$-2\lim_{n\to\infty}\left[\frac{\text{z}}{\sqrt{\overline{\text{q}}}}\cdot\int_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}\frac{1}{1-t^2}\space\text{d}t+\frac{\overline{\text{z}}}{\sqrt{\text{q}}}\cdot\int_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}\frac{1}{1-v^2}\space\text{d}v\right]=$$
Ahora, usa:
$$\int\frac{1}{1-x^2}\space\text{d}x=\frac{\ln\left|\frac{x+1}{x-1}\right|}{2}+\text{C}$$
$$-2\lim_{n\to\infty}\left[\frac{\text{z}}{\sqrt{\overline{\text{q}}}}\cdot\left[\frac{\ln\left|\frac{t+1}{t-1}\right|}{2}\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}+\frac{\overline{\text{z}}}{\sqrt{\text{q}}}\cdot\left[\frac{\ln\left|\frac{v+1}{v-1}\right|}{2}\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}\right]=$$ $$-2\lim_{n\to\infty}\left[\frac{\text{z}}{2\sqrt{\overline{\text{q}}}}\cdot\left[\ln\left|\frac{t+1}{t-1}\right|\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}+\frac{\overline{\text{z}}}{2\sqrt{\text{q}}}\cdot\left[\ln\left|\frac{v+1}{v-1}\right|\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}\right]=$$ $$-\lim_{n\to\infty}\left[\frac{\text{z}}{\sqrt{\overline{\text{q}}}}\cdot\left[\ln\left|\frac{t+1}{t-1}\right|\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}+\frac{\overline{\text{z}}}{\sqrt{\text{q}}}\cdot\left[\ln\left|\frac{v+1}{v-1}\right|\right]_{\frac{\sqrt{u}}{\sqrt{\text{q}}}}^{\frac{\sqrt{n}}{\sqrt{\text{q}}}}\right]=$$ $$-\lim_{n\to\infty}\left[\frac{\text{z}}{\sqrt{\overline{\text{q}}}}\cdot\left(\ln\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|-\ln\left|\frac{\frac{\sqrt{u}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{u}}{\sqrt{\text{q}}}-1}\right|\right)+\frac{\overline{\text{z}}}{\sqrt{\text{q}}}\cdot\left(\ln\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|-\ln\left|\frac{\frac{\sqrt{u}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{u}}{\sqrt{\text{q}}}-1}\right|\right)\right]=$$
Set:
$$\text{s}=\frac{\text{z}}{\sqrt{\overline{\text{q}}}}\to\overline{\text{s}}=\frac{\overline{\text{z}}}{\sqrt{\text{q}}}$$
$$-\lim_{n\to\infty}\left[\text{s}\cdot\left(\ln\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|-\ln\left|\frac{\frac{\sqrt{u}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{u}}{\sqrt{\text{q}}}-1}\right|\right)+\overline{\text{s}}\cdot\left(\ln\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|-\ln\left|\frac{\frac{\sqrt{u}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{u}}{\sqrt{\text{q}}}-1}\right|\right)\right]=$$
Supongamos que $u\space\wedge\space n\in\mathbb{R}^+$ :
- $$\text{A}=\left|\frac{\frac{\sqrt{u}}{\sqrt{\overline{\text{q}}}}+1}{\frac{\sqrt{u}}{\sqrt{\overline{\text{q}}}}-1}\right|=\left|\frac{\frac{\sqrt{u}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{u}}{\sqrt{\text{q}}}-1}\right|=\frac{\sqrt{\sqrt{2}+\sqrt{u}\cdot\sqrt{2(1+\sqrt{2})}+u}}{\sqrt{\sqrt{2}-\sqrt{u}\cdot\sqrt{2(1+\sqrt{2})}+u}}$$
- $$\text{B}=\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|=\left|\frac{\frac{\sqrt{n}}{\sqrt{\text{q}}}+1}{\frac{\sqrt{n}}{\sqrt{\text{q}}}-1}\right|=\frac{\sqrt{\sqrt{2}+\sqrt{n}\cdot\sqrt{2(1+\sqrt{2})}+n}}{\sqrt{\sqrt{2}-\sqrt{n}\cdot\sqrt{2(1+\sqrt{2})}+n}}$$
$$-\lim_{n\to\infty}\left[\text{s}\cdot\left(\ln\left(\text{B}\right)-\ln\left(\text{A}\right)\right)+\overline{\text{s}}\cdot\left(\ln\left(\text{B}\right)-\ln\left(\text{A}\right)\right)\right]=-\left(\text{s}+\overline{\text{s}}\right)\lim_{n\to\infty}\left[\ln\left(\text{B}\right)-\ln\left(\text{A}\right)\right]=$$
Aviso:
- $$\lim_{n\to\infty}\frac{\sqrt{\sqrt{2}+\sqrt{n}\cdot\sqrt{2(1+\sqrt{2})}+n}}{\sqrt{\sqrt{2}-\sqrt{n}\cdot\sqrt{2(1+\sqrt{2})}+n}}=1$$
-
$$\text{s}+\overline{\text{s}}=2\Re[\text{s}]=\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}$$
$$\left(\text{s}+\overline{\text{s}}\right)\ln\left(\text{A}\right)=\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}\cdot\ln\left(\text{A}\right)$$
