Question Number 196412 by Erico last updated on 25/Aug/23
$$\mathrm{Prove}\:\mathrm{that}\:\underset{\:{i}} {\int}^{\:+\infty} \left(\mathrm{cos2t}+\mathrm{isin2t}\right)\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}=\:\frac{\sqrt{\pi}}{\mathrm{2e}} \\ $$
Commented by JDamian last updated on 24/Aug/23
$${is}\:\boldsymbol{\mathrm{t}}\:{real}\:{or}\:{complex}? \\ $$
Answered by MrGaster last updated on 16/Feb/25
$$\int_{\mathrm{0}} ^{+\infty} \mathrm{cos2}{t}\centerdot{e}^{−{t}^{\mathrm{2}} } {dt}=\frac{\sqrt{\pi}}{\mathrm{2}{e}} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \mathrm{sin}\:\mathrm{2}{t}\centerdot{e}^{−{t}^{\mathrm{2}} } {dt}=\mathrm{0} \\ $$$$\underset{\:{i}} {\int}^{\:+\infty} \left(\mathrm{cos2t}+\mathrm{isin2t}\right)\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}=\:\frac{\sqrt{\pi}}{\mathrm{2e}} \\ $$