Question Number 61528 by maxmathsup by imad last updated on 04/Jun/19
$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:{cos}\left({zx}^{\mathrm{2}} \right){dx}\:{with}\:{z}\:\in\:{C}\:. \\ $$
Commented by maxmathsup by imad last updated on 04/Jun/19
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:{cos}\left({zx}^{\mathrm{2}} \right){dx}\:\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:{cos}\left({zx}^{\mathrm{2}} \right)\:\:{let}\:{z}\:={a}+{ib}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:{cos}\left(\left({a}+{ib}\right){x}^{\mathrm{2}} \right)\:\:\:{but}\:\:\:{cos}\left({z}\right)\:={ch}\left({iz}\right)\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:{ch}\left({i}\left({a}+{ib}\right){x}^{\mathrm{2}} \right){dx}\:=\int_{−\infty} ^{+\infty} \:{ch}\left({iax}^{\mathrm{2}} \:−{bx}^{\mathrm{2}} \right){dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{i}\left({iax}^{\mathrm{2}} −{bx}^{\mathrm{2}} \right)} \:+{e}^{−{i}\left({iax}^{\mathrm{2}} −{bx}^{\mathrm{2}} \right)} }{\mathrm{2}}\:{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{−{ax}^{\mathrm{2}} −{ibx}^{\mathrm{2}} } +{e}^{{ax}^{\mathrm{2}} +{ibx}^{\mathrm{2}} } }{\mathrm{2}}\:{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left({a}+{ib}\right){x}^{\mathrm{2}} } {dx}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left(−{a}−{ib}\right){x}^{\mathrm{2}} } {dx} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{e}^{−\left({a}+{ib}\right){x}^{\mathrm{2}} } {dx}=_{\sqrt{{a}+{ib}}{x}\:={t}} \:\:\:\:\:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \:\:\frac{{dt}}{\:\sqrt{{a}+{ib}}}\:=\frac{\pi}{\:\sqrt{{a}+{ib}}}\:=\frac{\pi}{\:\sqrt{{z}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:{e}^{−\left(−{a}−{ib}\right){x}^{\mathrm{2}} } {dx}\:=_{\sqrt{−{a}−{ib}}{x}\:={t}} \:\:\:\:\:\int_{−\infty} ^{+\infty} \:{e}^{−{t}^{\mathrm{2}} } \:\:\:\frac{{dt}}{\:\sqrt{−{a}−{ib}}}\:=\frac{\pi}{\:\sqrt{−{a}−{ib}}} \\ $$$$=\frac{\pi}{\:\sqrt{−\mathrm{1}}\sqrt{{a}+{ib}}}\:=\frac{\pi}{{i}\sqrt{{z}}}\:\Rightarrow\:\mathrm{2}{I}\:=\frac{\pi}{\mathrm{2}\sqrt{{z}}}\:+\frac{\pi}{\mathrm{2}{i}\sqrt{{z}}}\:=\frac{\pi}{\mathrm{2}\sqrt{{z}}}\:−\frac{{i}\pi}{\mathrm{2}\sqrt{{z}}}\:\:\Rightarrow{I}\:=\frac{\pi}{\mathrm{4}}\frac{\mathrm{1}−{i}}{\:\sqrt{{z}}} \\ $$$${if}\:{we}\:{take}\:{z}\:={re}^{{i}\theta} \:\:\:{we}\:{get}\:{I}\:=\frac{\pi}{\mathrm{4}}\:\frac{\mathrm{1}−{i}}{\:\sqrt{{r}}{e}^{{i}\frac{\theta}{\mathrm{2}}} }\:=\frac{\pi}{\mathrm{4}\sqrt{{r}}}\:\:\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:{e}^{−\frac{{i}\theta}{\mathrm{2}}} \:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}\sqrt{{r}}}\:{e}^{{i}\left(−\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)} \\ $$$${I}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}\sqrt{{r}}}\:{e}^{−{i}\left(\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}}\right)} \:\:. \\ $$
Commented by Smail last updated on 05/Jun/19
$${You}\:{mean}\:\:\int_{−\infty} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt}=\sqrt{\pi}\:\:\:{not}\:\pi \\ $$
Commented by maxmathsup by imad last updated on 08/Jun/19
$${yes}\:{sir}\:\:…. \\ $$
Commented by maxmathsup by imad last updated on 08/Jun/19
$${due}\:{to}\:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{t}^{\mathrm{2}} } {dt}\:=\sqrt{\pi}\:\:\:\:{the}\:{final}\:{answer}\:{is}\:{I}\:=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}\sqrt{{r}}}\:{e}^{−{i}\left(\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}}\right)} \\ $$$$ \\ $$