Ricevo da Francesca la seguente domanda:
Buongiorno Professore,
non so risolvere questo integrale improprio: \[\int\limits_{-1}^{+\infty }{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx\quad .\] Grazie.
Le rispondo così:
Cara Francesca,
essendo l’integrale improprio sia di prima che di seconda specie, suddividiamolo in due integrali: \[\int\limits_{-1}^{+\infty }{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx=\int\limits_{-1}^{0}{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx+\int\limits_{0}^{+\infty }{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx\] e calcoliamo, se esistono finiti, i seguenti limiti: \[\int\limits_{-1}^{0}{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx=\underset{\varepsilon \to 0}{\mathop{\lim }}\,\int\limits_{-1}^{\varepsilon }{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx\quad \quad \int\limits_{0}^{+\infty }{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx=\underset{h\to 0}{\mathop{\lim }}\,\left( \underset{k\to +\infty }{\mathop{\lim }}\,\int\limits_{h}^{k}{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx\quad \right).\]
Il primo integrale può essere così riformulato, essendo \(\sqrt{\left| x \right|}=\sqrt{-x}\) per ogni \(x<0\) e avendo posto \(t=-x\): \[\int\limits_{-1}^{\varepsilon }{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx=-\int\limits_{1}^{\varepsilon }{\frac{1}{\left( 2-t \right)\sqrt{t}}}dt=\int\limits_{\varepsilon }^{1}{\frac{1}{\left( 2-t \right)\sqrt{t}}}dt\] e poiché, posto \(\sqrt{t}=p,\ dt=2pdp\):\[\int{\frac{1}{\left( 2-t \right)\sqrt{t}}dt}=2\int{\frac{1}{2-{{p}^{2}}}dp=\frac{\sqrt{2}}{2}}\int{\frac{1}{p+\sqrt{2}}dp}-\frac{\sqrt{2}}{2}\int{\frac{1}{p-\sqrt{2}}dp}=\frac{\sqrt{2}}{2}\ln \left| \frac{\sqrt{t}+\sqrt{2}}{\sqrt{t}-\sqrt{2}} \right|+c\] si ha il seguente risultato:\[\int\limits_{-1}^{0}{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx=\underset{\varepsilon \to 0}{\mathop{\lim }}\,\left[ \frac{\sqrt{2}}{2}\ln \left| \frac{\sqrt{t}+\sqrt{2}}{\sqrt{t}-\sqrt{2}} \right| \right]_{\varepsilon }^{1}=\frac{\sqrt{2}}{2}\left( \ln \left| \frac{1+\sqrt{2}}{1-\sqrt{2}} \right|-\underset{\varepsilon \to 0}{\mathop{\lim }}\,\ln \left| \frac{\sqrt{\varepsilon }+\sqrt{2}}{\sqrt{\varepsilon }-\sqrt{2}} \right| \right)=\sqrt{2}\ln \left( 1+\sqrt{2} \right)\quad .\] Poiché per ogni \(x>0\) si ha \(\sqrt{\left| x \right|}=\sqrt{x}\), il secondo integrale risulta: \[\underset{h\to 0}{\mathop{\lim }}\,\left( \underset{k\to +\infty }{\mathop{\lim }}\,\int\limits_{h}^{k}{\frac{1}{\left( x+2 \right)\sqrt{x}}}dx \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \underset{k\to +\infty }{\mathop{\lim }}\,\int\limits_{\sqrt{h}}^{\sqrt{k}}{\frac{2}{{{p}^{2}}+2}}dp \right)=\sqrt{2}\underset{h\to 0}{\mathop{\lim }}\,\left( \underset{k\to +\infty }{\mathop{\lim }}\,\int\limits_{\sqrt{h}}^{\sqrt{k}}{\frac{1/\sqrt{2}}{1+{{\left( p/\sqrt{2} \right)}^{2}}}}dp \right)=\] \[=\sqrt{2}\underset{h\to 0}{\mathop{\lim }}\,\left( \underset{k\to +\infty }{\mathop{\lim }}\,\left[ \arctan \left( \frac{\sqrt{x}}{\sqrt{2}} \right) \right]_{h}^{k} \right)=\sqrt{2}\left( \underset{k\to +\infty }{\mathop{\lim }}\,\arctan \left( \frac{\sqrt{k}}{\sqrt{2}} \right)-\underset{h\to 0}{\mathop{\lim }}\,\arctan \left( \frac{\sqrt{h}}{\sqrt{2}} \right) \right)=\frac{\pi \sqrt{2}}{2}\quad .\] In conclusione, possiamo dire che: \[\int\limits_{-1}^{+\infty }{\frac{1}{\left( x+2 \right)\sqrt{\left| x \right|}}}dx=\sqrt{2}\ln \left( 1+\sqrt{2} \right)+\frac{\pi \sqrt{2}}{2}\approx 3,468\quad .\]
Massimo Bergamini