Integration Questions

Question Number 45841 by Rauny last updated on 17/Oct/18

$$\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{ix}^{\mathrm{2}} } {dx}=?? \\$$$$\mathrm{plz}.. \\$$

Commented by MJS last updated on 17/Oct/18

$$\frac{\sqrt{\left(\mathrm{2}\pi\right)}}{\mathrm{4}}−\frac{\sqrt{\left(\mathrm{2}\pi\right)}}{\mathrm{4}}\mathrm{i} \\$$

Commented by maxmathsup by imad last updated on 17/Oct/18

$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{ix}^{\mathrm{2}} } {dx}\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({x}\sqrt{{i}}\right)^{\mathrm{2}} } {dx}\:\:{changement}\:{x}\sqrt{{i}}={t}\:{give} \\$$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{ix}^{\mathrm{2}} } {dx}\:=\frac{\mathrm{1}}{\sqrt{{i}}}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:=\frac{\mathrm{1}}{\sqrt{{i}}}\:\frac{\sqrt{\pi}}{\mathrm{2}}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:\frac{\mathrm{1}}{{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\$$$$=\frac{\sqrt{\pi}}{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:−\frac{{i}}{\sqrt{\mathrm{2}}}\right)\:=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\:−{i}\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}}\:−{i}\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}} \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Oct/18

$${t}={ix}^{\mathrm{2}} \:{dt}=\mathrm{2}{ixdx} \\$$$${dx}=\frac{{dt}}{\mathrm{2}{ix}}=\frac{{dt}}{\mathrm{2}{i}×\sqrt{{t}}}×\sqrt{{i}} \\$$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} ×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}} \\$$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dt}\:\:\leftarrow{gamma}\:{function} \\$$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}}×\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\$$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{{i}}}×\sqrt{\pi}\:\:\:\:\:\:\:\:\left\{{since}\:\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}\:\right\} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{{i}}\:}{{i}}×\sqrt{\pi}\: \\$$$${now}\:{calculation}\:{of}\:\sqrt{{i}}\: \\$$$$=\frac{\mathrm{1}}{\sqrt{}\mathrm{2}}×\sqrt{\mathrm{1}−\mathrm{1}+\mathrm{2}{i}}\: \\$$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}×\sqrt{\mathrm{1}^{\mathrm{2}} +{i}^{\mathrm{2}} +\mathrm{2}×{i}×\mathrm{1}}\: \\$$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}×\sqrt{\left(\mathrm{1}+{i}\right)^{\mathrm{2}} }\: \\$$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}×\left(\mathrm{1}+{i}\right) \\$$$${so}\:{the}\:{ans}\:{is} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}×\frac{\mathrm{1}+{i}}{{i}}×\sqrt{\pi}\: \\$$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}×\left(\mathrm{1}+\frac{\mathrm{1}}{{i}}\right)×\sqrt{\pi}\: \\$$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}×\left(\mathrm{1}−{i}\right)×\sqrt{\pi}\: \\$$$$\\$$$$\left\{\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} ×{y}^{{n}−\mathrm{1}} {dy}=\lceil\left({n}\right)\right\} \\$$