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Question Number 193231 by C2coder last updated on 07/Jun/23

Answered by leodera last updated on 08/Jun/23

$$F\left(a\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{-x^2}\sin {}^2\left({ax}^2\right)}{x^2}dx$$$$F^{\prime }\left(a\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}\sin \left(2{ax}^2\right)dx$$$$F^{\prime }\left(a\right)=2{\int }_0^{\mathrm{\infty }}{e}^{-x^2}\sin \left(2{ax}^2\right)dx$$$$$$$$letu=2{ax}^2\Rightarrow \frac{1}{2\sqrt{2a}}{u}^{-\frac{1}{2}}du=dx$$$$F^{\prime }\left(a\right)=\frac{1}{\sqrt{2a}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{u}^{-\frac{1}{2}}{e}^{-\frac{1}{2a}u}\sin \left(u\right)du$$$$F^{\prime }\left(a\right)=Im\frac{1}{\sqrt{2a}}{\int }_0^{\mathrm{\infty }}{u}^{-\frac{1}{2}}{e}^{-(\frac{1}{2}-i)u}du$$$$F^{\prime }\left(a\right)=Im\frac{1}{\sqrt{2a}}\left\{\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{{(\frac{1}{2}-i)}^{\frac{1}{2}}}\right\}$$$$F^{\prime }\left(a\right)=Im\sqrt{\frac{\pi }{2a}}\left\{\frac{1}{{\left(\frac{\sqrt{5}}{2}\right)}^{\frac{1}{2}}{e}^{-i\frac{{\tan }^{-1}\left(2\right)}{2}}}\right\}$$$$F^{\prime }\left(a\right)=-\sqrt{\frac{\pi }{a\sqrt{5}}}\sin \left(\frac{{\tan }^{-1}\left(2\right)}{2}\right)$$$$F\left(a\right)=2\sqrt{\frac{a\pi }{\sqrt{5}}}\sin \left(\frac{{\tan }^{-1}2}{2}\right)+C$$$$F\left(0\right)=0$$$$soC=0$$$$F\left(1\right)=2\sqrt{\frac{\pi }{\sqrt{5}}}\sin \left(\frac{{\tan }^{-1}\left(2\right)}{2}\right)$$$$$$

Commented by C2coder last updated on 08/Jun/23

$$thanksalotman$$$$$$

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