calculte-sin-pix-2-x-2-2x-2-2-dx-

Question Number 137536 by Mathspace last updated on 03/Apr/21

c a l c u l t e \int_{- \infty}^{\infty} \frac{s i n (π x^{2})}{{(x^{2} + 2 x + 2)}^{2}} d x

Answered by mathmax by abdo last updated on 04/Apr/21

let f (a) = \int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{x^{2} + 2 x + a} dx with a > 1

we have f^{^{'}} (a) = - \int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{{(x^{2} + 2 x + a)}^{2}} \Rightarrow f^{^{'}} (2) = - \int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{{(x^{2} + 2 x + 2)}^{2}} \Rightarrow

\int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{{(x^{2} + 2 x + 2)}^{2}} = - f^{^{'}} (2)

we have f (a) = Im (\int_{- \infty}^{+ \infty} \frac{e^{i π x^{2}}}{x^{2} + 2 x + a} dx) let φ (z) = \frac{e^{i π z^{2}}}{z^{2} + 2 z + a}

poles of φ ? Δ^{^{'}} = 1 - a < 0 \Rightarrow z_{1} = - 1 + i \sqrt{a - 1} and z_{2} = - 1 - i \sqrt{a - 1}

φ (z) = \frac{e^{i π z^{2}}}{(z - z_{1}) (z - z_{2})} residus theorem \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, z_{1}) = 2 i π \times \frac{e^{i π z_{1}^{2}}}{2 i \sqrt{a - 1}}

= \frac{π}{\sqrt{a - 1}} e^{i π {1 - 2 i \sqrt{a - 1} - a + 1)} = \frac{π}{\sqrt{a - 1}} e^{i π {2 - a - 2 i \sqrt{a - 1}}}

= \frac{π}{\sqrt{a - 1}} e^{i π (2 - a)} . e^{2 π \sqrt{a - 1}} = \frac{π e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}} {\cos (π (2 - a)) + isin (π (2 - a))} \Rightarrow

f (a) = - \frac{π e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}} . \sin (π a) \Rightarrow

- f^{^{'}} (a) = π {(\frac{e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}})}^{,} \sin (π a) + π^{2} \cos (π a) . \frac{e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}} but

{(\frac{e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}})}^{^{'}} = \frac{\frac{2 π}{2 \sqrt{a - 1}} e^{2 π \sqrt{a - 1}} . \sqrt{a - 1} - e^{2 π \sqrt{a - 1}} . \frac{1}{2 \sqrt{a - 1}}}{a - 1}

= \frac{π e^{2 π \sqrt{a - 1}} - \frac{1}{2 \sqrt{a - 1}} e^{2 π \sqrt{a - 1}}}{a - 1} = \frac{(2 π \sqrt{a - 1} - 1) e^{2 π \sqrt{a - 1}}}{2 (a - 1) \sqrt{a - 1}} \Rightarrow

f^{^{'}} (a) = - \frac{π}{2} \frac{(2 π \sqrt{a - 1} - 1) e^{2 π \sqrt{a - 1}}}{(a - 1) \sqrt{a - 1}} \sin (π a) - π^{2} \cos (π a) . \frac{e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}}

\Rightarrow f^{^{'}} (2) = - \frac{π}{2} \frac{(2 π - 1) e^{2 π}}{1} . 0 - π^{2} e^{2 π} = - π^{2} e^{2 π} \Rightarrow

\int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{{(x^{2} + 2 x + 2)}^{2}} = π^{2} e^{2 π}