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Question Number 6775 by Tawakalitu. last updated on 24/Jul/16

$$\int_{\mathrm{0}} ^{\infty} {e}^{{t}^{\mathrm{2}} } {e}^{−{st}} \:\:{dt} \\$$$$\\$$$${please}\:{help} \\$$

Answered by Yozzii last updated on 24/Jul/16

$${b}=\int_{\mathrm{0}} ^{\infty} {e}^{{t}^{\mathrm{2}} −{st}} {dt}=\int_{\mathrm{0}} ^{\infty} {e}^{\left({t}−\mathrm{0}.\mathrm{5}{s}\right)^{\mathrm{2}} −\mathrm{0}.\mathrm{25}{s}^{\mathrm{2}} } {dt} \\$$$${b}={e}^{−\mathrm{0}.\mathrm{25}{s}^{\mathrm{2}} } \int_{\mathrm{0}} ^{\infty} {e}^{\left({t}−\mathrm{0}.\mathrm{5}{s}\right)^{\mathrm{2}} } {dt}. \\$$$${u}={t}−\mathrm{0}.\mathrm{5}{s}\Rightarrow{du}={dt} \\$$$$\therefore{b}={e}^{−\mathrm{0}.\mathrm{25}{s}^{\mathrm{2}} } \left(\int_{−\mathrm{0}.\mathrm{5}{s}} ^{\mathrm{0}} {e}^{{u}^{\mathrm{2}} } {du}+\int_{\mathrm{0}} ^{\infty} {e}^{{u}^{\mathrm{2}} } {du}\right) \\$$$$\int_{\mathrm{0}} ^{\infty} {e}^{{u}^{\mathrm{2}} } {du}\:{is}\:{divergent}\:{while}\:\int_{−\mathrm{0}.\mathrm{5}{s}} ^{\mathrm{0}} {e}^{{u}^{\mathrm{2}} } {du} \\$$$${is}\:{finite}. \\$$$$\therefore\:{For}\:{nonzero}\:{s}\in\mathbb{R},\:{b}={e}^{−\mathrm{0}.\mathrm{25}{s}^{\mathrm{2}} } \left\{\int_{−\mathrm{0}.\mathrm{5}{s}} ^{\mathrm{0}} {e}^{{u}^{\mathrm{2}} } {du}+\int_{\mathrm{0}} ^{\infty} {e}^{{u}^{\mathrm{2}} } {du}\right\} \\$$$${does}\:{not}\:{exist}.\: \\$$$${If}\:{s}=\mathrm{0},{b}=\int_{\mathrm{0}} ^{\infty} {e}^{{t}^{\mathrm{2}} } {dt}\:{which}\:{is}\:{divergent}\:{also}. \\$$

Commented by Tawakalitu. last updated on 24/Jul/16

$${Thanks}\:{so}\:{much},\:{That}\:{means}\:{the}\:{queastion}\:{is}\:{wrong}\:?? \\$$

Commented by Yozzii last updated on 24/Jul/16

$${Not}\:{necessarily}.\:{The}\:{question}\:{may}\:{have} \\$$$${asked}\:{to}\:{see}\:{whether}\:{the}\:{improper} \\$$$${integral}\:{is}\:{convergent}\:{or}\:{divergent}. \\$$

Commented by Tawakalitu. last updated on 24/Jul/16

$${Thanks} \\$$

Commented by Tawakalitu. last updated on 25/Jul/16

$${The}\:{question}\:{is}\:{evaluate} \\$$