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Question Number 49232 by Abdo msup. last updated on 04/Dec/18

$$letf\left(a\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}cos(x^2+ax+1)dx$$$$1)calculatef\left(a\right)and{f}^{{}^{\prime }}\left(a\right)$$$$2)find{f}^{\left(n\right)}\left(a\right)$$

Commented by Abdo msup. last updated on 06/Dec/18

$$letg\left(a\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}sin(x^2+ax+1)dxwehsve$$$$f\left(a\right)-ig\left(a\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-i(x^2+ax+1)}dx$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-i(x^2+2\frac{a}{2}x+\frac{a^2}{4}+1-\frac{a^2}{4})}dx$$$$=e^{-i(1-\frac{a^2}{4})}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-{\left(\sqrt{i}(x+\frac{a}{2})\right)}^2}dx$$$${=}_{\sqrt{i}(x+\frac{a}{2})=u}{e}^{-i(1-\frac{a^2}{4})}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-u^2}\frac{du}{\sqrt{i}}$$$$=\sqrt{\pi }{e}^{-i(1-\frac{a^2}{4})}{e}^{-\frac{i\pi }{4}}=\sqrt{\pi }{e}^{-i(1+\frac{\pi }{4}-\frac{a^2}{2})}$$$$\sqrt{\pi }\{cos(1+\frac{\pi }{4}-\frac{a^2}{4})-isin(1+\frac{\pi }{4}-\frac{a^2}{4})\}\Rightarrow$$$$f\left(a\right)=\sqrt{\pi }cos(1+\frac{\pi }{4}-\frac{a^2}{4})alsowehave$$$$g\left(a\right)=\sqrt{\pi }sin(1+\frac{\pi }{4}-\frac{a^2}{4}).$$

Commented by Abdo msup. last updated on 06/Dec/18

$$f^{{}^{\prime }}\left(a\right)=-\sqrt{\pi }(-\frac{a}{2})sin(1+\frac{\pi }{4}-\frac{a^2}{4})$$$$=\frac{a\sqrt{\pi }}{2}sin(1+\frac{\pi }{4}-\frac{a^2}{4}).$$

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