Question Number 219349 by SdC355 last updated on 23/Apr/25
$$\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{r}^{\mathrm{2}} } \mathrm{cos}\left({r}\right)\:\mathrm{d}{r}=?? \\ $$
Answered by vnm last updated on 23/Apr/25
![∫_0 ^∞ e^(−(x+a)^2 ) dx+∫_0 ^∞ e^(−(x−a)^2 ) dx= ∫_0 ^∞ e^(−(x+a)^2 ) d(x+a)+∫_0 ^∞ e^(−(x−a)^2 ) d(x−a)= [((t=x+a)),((u=a−x)) ]=∫_a ^∞ e^(−t^2 ) dt+∫_a ^(−∞) e^(−(−u)^2 ) d(−u)= ∫_(−∞) ^a e^(−u^2 ) du+∫_a ^∞ e^(−t^2 ) dt=∫_(−∞) ^∞ e^(−u^2 ) du=(√π) ∫_0 ^∞ e^(−r^2 ) cosr∙dr=(1/2)∫_0 ^∞ e^(−r^2 ) (e^(ir) +e^(−ir) )dr= (1/2)∫_0 ^∞ (e^(−(1/4)) e^(−r^2 +2(i/2)r−((i/2))^2 ) + e^(−(1/4)) e^(−r^2 −2(i/2)r−((i/2))^2 ) )dr= (1/2)e^(−(1/4)) (∫_0 ^∞ e^(−(r+(i/2))^2 ) dr+∫_0 ^∞ e^(−(r−(i/2))^2 ) dr)= (1/2)e^(−(1/4)) (∫_(i/2) ^∞ e^(−t^2 ) dt+∫_(−∞) ^(i/2) e^(−t^2 ) dt)=(1/2)e^(−(1/4)) ∫_(−∞) ^∞ e^(−t^2 ) dt=(1/2)e^(−(1/4)) (√π)](Q219368.png)
$$\int_{\mathrm{0}} ^{\infty} {e}^{−\left({x}+{a}\right)^{\mathrm{2}} } {dx}+\int_{\mathrm{0}} ^{\infty} {e}^{−\left({x}−{a}\right)^{\mathrm{2}} } {dx}= \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−\left({x}+{a}\right)^{\mathrm{2}} } {d}\left({x}+{a}\right)+\int_{\mathrm{0}} ^{\infty} {e}^{−\left({x}−{a}\right)^{\mathrm{2}} } {d}\left({x}−{a}\right)=\begin{bmatrix}{{t}={x}+{a}}\\{{u}={a}−{x}}\end{bmatrix}=\int_{{a}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt}+\int_{{a}} ^{−\infty} {e}^{−\left(−{u}\right)^{\mathrm{2}} } {d}\left(−{u}\right)= \\ $$$$\int_{−\infty} ^{{a}} {e}^{−{u}^{\mathrm{2}} } {du}+\int_{{a}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt}=\int_{−\infty} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du}=\sqrt{\pi} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{r}^{\mathrm{2}} } \mathrm{cos}{r}\centerdot{dr}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{r}^{\mathrm{2}} } \left({e}^{{ir}} +{e}^{−{ir}} \right){dr}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left({e}^{−\frac{\mathrm{1}}{\mathrm{4}}} {e}^{−{r}^{\mathrm{2}} +\mathrm{2}\frac{{i}}{\mathrm{2}}{r}−\left(\frac{{i}}{\mathrm{2}}\right)^{\mathrm{2}} } +\:\:{e}^{−\frac{\mathrm{1}}{\mathrm{4}}} {e}^{−{r}^{\mathrm{2}} −\mathrm{2}\frac{{i}}{\mathrm{2}}{r}−\left(\frac{{i}}{\mathrm{2}}\right)^{\mathrm{2}} } \right){dr}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\mathrm{1}}{\mathrm{4}}} \left(\int_{\mathrm{0}} ^{\infty} {e}^{−\left({r}+\frac{{i}}{\mathrm{2}}\right)^{\mathrm{2}} } {dr}+\int_{\mathrm{0}} ^{\infty} {e}^{−\left({r}−\frac{{i}}{\mathrm{2}}\right)^{\mathrm{2}} \:} {dr}\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\mathrm{1}}{\mathrm{4}}} \left(\int_{\frac{{i}}{\mathrm{2}}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt}+\int_{−\infty} ^{\frac{{i}}{\mathrm{2}}} {e}^{−{t}^{\mathrm{2}} } {dt}\right)=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\mathrm{1}}{\mathrm{4}}} \int_{−\infty} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\mathrm{1}}{\mathrm{4}}} \sqrt{\pi} \\ $$