Integral von $$$2^{x^{2}}$$$

Bestimme $$$\int 2^{x^{2}}\, dx$$$.

Basis ändern:

$${\color{red}{\int{2^{x^{2}} d x}}} = {\color{red}{\int{e^{x^{2} \ln{\left(2 \right)}} d x}}}$$

Sei $$$u=x \sqrt{\ln{\left(2 \right)}}$$$.

Dann $$$du=\left(x \sqrt{\ln{\left(2 \right)}}\right)^{\prime }dx = \sqrt{\ln{\left(2 \right)}} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{\sqrt{\ln{\left(2 \right)}}}$$$.

Das Integral lässt sich umschreiben als

$${\color{red}{\int{e^{x^{2} \ln{\left(2 \right)}} d x}}} = {\color{red}{\int{\frac{e^{u^{2}}}{\sqrt{\ln{\left(2 \right)}}} d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{\sqrt{\ln{\left(2 \right)}}}$$$ und $$$f{\left(u \right)} = e^{u^{2}}$$$ an:

$${\color{red}{\int{\frac{e^{u^{2}}}{\sqrt{\ln{\left(2 \right)}}} d u}}} = {\color{red}{\frac{\int{e^{u^{2}} d u}}{\sqrt{\ln{\left(2 \right)}}}}}$$

Dieses Integral (Imaginäre Fehlerfunktion) besitzt keine geschlossene Form:

$$\frac{{\color{red}{\int{e^{u^{2}} d u}}}}{\sqrt{\ln{\left(2 \right)}}} = \frac{{\color{red}{\left(\frac{\sqrt{\pi} \operatorname{erfi}{\left(u \right)}}{2}\right)}}}{\sqrt{\ln{\left(2 \right)}}}$$

Zur Erinnerung: $$$u=x \sqrt{\ln{\left(2 \right)}}$$$:

$$\frac{\sqrt{\pi} \operatorname{erfi}{\left({\color{red}{u}} \right)}}{2 \sqrt{\ln{\left(2 \right)}}} = \frac{\sqrt{\pi} \operatorname{erfi}{\left({\color{red}{x \sqrt{\ln{\left(2 \right)}}}} \right)}}{2 \sqrt{\ln{\left(2 \right)}}}$$

Daher,

$$\int{2^{x^{2}} d x} = \frac{\sqrt{\pi} \operatorname{erfi}{\left(x \sqrt{\ln{\left(2 \right)}} \right)}}{2 \sqrt{\ln{\left(2 \right)}}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{2^{x^{2}} d x} = \frac{\sqrt{\pi} \operatorname{erfi}{\left(x \sqrt{\ln{\left(2 \right)}} \right)}}{2 \sqrt{\ln{\left(2 \right)}}}+C$$

$$$\int 2^{x^{2}}\, dx = \frac{\sqrt{\pi} \operatorname{erfi}{\left(x \sqrt{\ln\left(2\right)} \right)}}{2 \sqrt{\ln\left(2\right)}} + C$$$A