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Question Number 184027 by SEKRET last updated on 02/Jan/23

$$\:\:\: \\$$$$\\$$$$\\$$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\boldsymbol{\mathrm{e}}^{β\boldsymbol{{tg}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)} \:\boldsymbol{{d}\mathrm{x}}\:=\:??? \\$$$$\\$$$$\\$$$$\\$$$$\\$$

Answered by witcher3 last updated on 03/Jan/23

$$=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{β{x}^{\mathrm{2}} } }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\leqslant\int_{\mathrm{0}} ^{\infty} {e}^{β{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}..{cv} \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{β{ax}^{\mathrm{2}} } }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}={F}\left({a}\right) \\$$$${f}'\left({a}\right)β{f}\left({a}\right)=β\int_{\mathrm{0}} ^{\infty} {e}^{β{ax}^{\mathrm{2}} } {dx} \\$$$$=\int_{\mathrm{0}} ^{\infty} {e}^{β\left({x}\sqrt{{a}}\right)^{\mathrm{2}} } .{dx}=β\frac{\sqrt{\pi}}{\:\mathrm{2}\sqrt{{a}}} \\$$$${f}'β{f}=β\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{{a}}} \\$$$${f}\left({a}\right)={ke}^{{a}} \\$$$${k}'=β\frac{\sqrt{\pi}}{\mathrm{2}}\frac{{e}^{β{a}} }{\:\sqrt{{a}}}\Rightarrow{k}=β\frac{\sqrt{\pi}}{\mathrm{2}}\int\:\frac{{e}^{β{a}} }{\:\sqrt{{a}}}{da}=β\sqrt{\pi}\int{e}^{β{t}^{\mathrm{2}} } {dt} \\$$$$=\frac{\pi}{\mathrm{2}}{erfc}\left({t}\right)+{c},{c}\in\mathbb{R} \\$$$${f}\left({a}\right)=\frac{\pi}{\mathrm{2}}{erfc}\left(\sqrt{{a}}\right){e}^{β{a}} +{ce}^{β{a}} \\$$$${f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}\Leftrightarrow{c}=\mathrm{0} \\$$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{β{atg}^{\mathrm{2}} \left({x}\right)} {dx}=\frac{\pi}{\mathrm{2}}{erfc}\left(\sqrt{{a}}\right){e}^{β{a}} \\$$$${f}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{β{tg}^{\mathrm{2}} \left({x}\right)} {dx}=\frac{\pi}{\mathrm{2}{e}}{erfc}\left(\mathrm{1}\right) \\$$

Commented by SEKRET last updated on 05/Jan/23

$$\boldsymbol{\mathrm{thanks}}\:\:\boldsymbol{\mathrm{sir}}.\:\boldsymbol{\mathrm{beatiful}}\:\boldsymbol{\mathrm{solution}} \\$$

Commented by witcher3 last updated on 05/Jan/23

$${withe}\:{pleasur} \\$$