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Question Number 153949 by talminator2856791 last updated on 12/Sep/21

$$$$$${\int }_0^{\mathrm{\infty }}\sin \left(x^2\right)\cos \left(x^3\right)dx$$$$$$

Answered by mindispower last updated on 16/Sep/21

$$=Im{\int }_0^{\mathrm{\infty }}{e}^{{ix}^2}cos\left(x^3\right)dx$$$$A={\int }_0^{\mathrm{\infty }}{e}^{-{ix}^2}cos\left(x^3\right)dx$$$$Ai\left(z\right)=\frac{e^{-\frac{2}{3}{z}^{\frac{3}{2}}}}{\pi }{\int }_0^{\mathrm{\infty }}{e}^{-\sqrt{z}{t}^2}cos\left(\frac{t^3}{3}\right)dt...1$$$$AiryFunctionintegralrepresentation$$$$t\Rightarrow y.{\left(3\right)}^{\frac{1}{3}}$$$$Ai\left(z\right)=k\left(z\right){\int }_0^{\mathrm{\infty }}{e}^{-\sqrt{z}.y^2{\left(3\right)}^{\frac{2}{3}}}.3^{\frac{1}{3}}cos4\left(y^3\right)d$$$$Ai(-\frac{1}{3^{\frac{4}{3}}})=K(-\frac{1}{3^{\frac{4}{3}}})3^{\frac{1}{3}}{\int }_0^{\mathrm{\infty }}{e}^{-{iy}^2}cos\left(y^3\right)dy$$$$\frac{e^{-\frac{2}{27}.e^{\frac{3\pi i}{2}}}}{\pi }A$$$$A=\pi e^{-\frac{2}{27}i}.\frac{Ai(-\frac{1}{3^{\frac{4}{3}}})}{\sqrt[3]{3}}$$$${\int }_0^{\mathrm{\infty }}sin\left(x^2\right)cos\left(x^3\right)dx=-ImA$$$$=-\frac{\pi }{\sqrt[3]{3}}sin(-\frac{2}{27})Ai(-\frac{1}{3^{\frac{4}{3}}})$$$$=\frac{\pi sin\left(\frac{2}{27}\right)A_i(-\frac{1}{3\sqrt[3]{3}})}{\sqrt[3]{3}}$$$$wherre{A}_i...Airyfunction$$$${\int }_0^{\mathrm{\infty }}sin\left(x^2\right)cos\left(x^3\right)dx=\frac{\pi sin\left(\frac{2}{27}\right)A_i(-\frac{1}{3\sqrt[3]{3}})}{\sqrt[3]{3}}$$$$$$$$$$

Commented by talminator2856791 last updated on 23/Sep/21

$$\mathrm{integral}\mathrm{master}!$$

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