Question Number 219268 by MrGaster last updated on 21/Apr/25
$${f}\left({x},{y}\right)=\mathrm{ln}\int_{\mathrm{0}} ^{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } {e}^{{t}^{\mathrm{2}} } {dt},{f}\left({x}\right)\left(\mathrm{1},\mathrm{2}\right)=? \\ $$
Answered by zetamaths last updated on 21/Apr/25
![exp(f(x;y))=∫_0 ^5 e^(−(−x^2 )) dx=∫_A e^(−(ix)^2 ) dx A=[0;5] let u=ix x=((1−x)/(1×i×i))u x=−iu dx=−idu ∫_A e^(−u^2 ) du×(−1)=−i((√π)/2)erf(u)=−i((√π)/2)erf(ix) [((erfi(x)×(√(π×(1/2^2 ))))) ]_0 ^5 =(erfi(5)×((√π)/2))−(erfi(0)×(√(π(1/4))))=erfi(5)×((√π)/2) f(1;2)^ =ln(erfi(5))+ln(((√π)/2))=_(afteer computing) 21.6](https://www.tinkutara.com/question/Q219269.png)
$${exp}\left({f}\left({x};{y}\right)\right)=\int_{\mathrm{0}} ^{\mathrm{5}} {e}^{−\left(−{x}^{\mathrm{2}} \right)} {dx}=\int_{\mathbb{A}} {e}^{−\left({ix}\right)^{\mathrm{2}} } {dx}\:\:\mathbb{A}=\left[\mathrm{0};\mathrm{5}\right] \\ $$$${let}\:{u}={ix} \\ $$$${x}=\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}×{i}×{i}}{u} \\ $$$${x}=−{iu} \\ $$$${dx}=−{idu} \\ $$$$\int_{\mathbb{A}} {e}^{−{u}^{\mathrm{2}} } {du}×\left(−\mathrm{1}\right)=−{i}\frac{\sqrt{\pi}}{\mathrm{2}}{erf}\left({u}\right)=−{i}\frac{\sqrt{\pi}}{\mathrm{2}}{erf}\left({ix}\right)\begin{bmatrix}{{erfi}\left({x}\right)×\sqrt{\pi×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }}}\end{bmatrix}_{\mathrm{0}} ^{\mathrm{5}} =\left({erfi}\left(\mathrm{5}\right)×\frac{\sqrt{\pi}}{\mathrm{2}}\right)−\left({erfi}\left(\mathrm{0}\right)×\sqrt{\pi\frac{\mathrm{1}}{\mathrm{4}}}\right)={erfi}\left(\mathrm{5}\right)×\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$${f}\left(\mathrm{1};\mathrm{2}\overset{} {\right)}={ln}\left({erfi}\left(\mathrm{5}\right)\right)+{ln}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right)\underset{{afteer}\:{computing}} {=}\mathrm{21}.\mathrm{6} \\ $$$$ \\ $$