Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

∫_0 ^( ∞) e^(−ix^2 ) dx=?? plz..

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

Commented by MJS last updated on 17/Oct/18

((√((2π)))/4)−((√((2π)))/4)i

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

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

∫_0 ^∞ e^(−ix^2 ) dx =∫_0 ^∞ e^(−(x(√i))^2 ) dx changement x(√i)=t give ∫_0 ^∞ e^(−ix^2 ) dx =(1/(√i)) ∫_0 ^∞ e^(−t^2 ) dt =(1/(√i)) ((√π)/2) =((√π)/2) (1/e^(i(π/4)) ) =((√π)/2) e^(−((iπ)/4)) =((√π)/2) ((1/(√2)) −(i/(√2))) =((√π)/(2(√2))) −i((√π)/(2(√2))) =((√(2π))/4) −i((√(2π))/4)

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

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

t=ix^2 dt=2ixdx dx=(dt/(2ix))=(dt/(2i×(√t)))×(√i) ∫_0 ^∞ e^(−t) ×t^(−(1/2)) ×(1/(2(√i))) (1/(2(√i)))∫_0 ^∞ e^(−t) ×t^((1/2)−1) dt ←gamma function (1/(2(√i)))×⌈((1/2)) (1/(2(√i)))×(√π) {since ⌈((1/2))=(√π) } =(1/2)×(((√i) )/i)×(√π) now calculation of (√i) =(1/((√)2))×(√(1−1+2i)) =(1/(√2))×(√(1^2 +i^2 +2×i×1)) =(1/(√2))×(√((1+i)^2 )) =(1/(√2))×(1+i) so the ans is =(1/2)×(1/((√2) ))×((1+i)/i)×(√π) =(1/(2(√2) ))×(1+(1/i))×(√π) =(1/(2(√2)))×(1−i)×(√π) {∫_0 ^∞ e^(−y) ×y^(n−1) dy=⌈(n)}

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com