How to integrate sin(x^2)

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How to integrate sin(x^2)

Check out this answer for a real method to evaluate this integral. First note that $$ \int_0^\infty e^{-x^2}\,\mathrm{d}x=\frac{\sqrt\pi}2\tag{1} $$ Integral $$ \int_\Gamma e^{-z^2}\,\mathrm{d}z=0\tag{2} $$ Move to the $\gamma$ curve that consists of the line from $0$0 to $$R$ and then counterclockwise along the circular arc from $$R to $Re^{i\pi/4}$ and after Then back down the line from $Re^{i\pi/4}$ to $0$ must be $0$ because $e^{-z^2}$ has no uniqueness within $\gamma$. The following note is that $\int_\gamma e^{-z^2}\,\mathrm{d}z$ down the line from $0$ to $$R$ as $R\to\infty$ tends to $(1)$. The following note is that $\int_\gamma e^{-z^2}\mathrm{d}z$ along the arc of the radius circle $R$ from $$R to $Re^{i\pi/4}$ when $$R\to\in\infty$ moves to zero. Finally, note that $\int_\gamma e^{-z^2}\mathrm{d}z$ down the line from $Re^{i\pi/4}$ to $0$ is $$ -e^{i\pi/4}\int_0^\infty e^{-ix^2}\,\mathrm{d}x\tag{3} $$ Therefore, $(1)$ plus $(3)$ is $0$, so we get $$ \int_0^\infty e^{-ix^2}\,\mathrm{d}x=\frac{1-i} {\sqrt2}\frac{\sqrt\pi}2\tag{4} $$ Takes the imaginary part of $(4)$ returns that $$ \int_0^\infty\sin(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{5} $$ Or $$ \int_{-\infty}^^\infty\sin(x^2)\,\mathrm{d}x=\sqrt{\frac\pi2}\tag{6} $$ $\begingroup$ exchange $u not working. I don't see anything to do with replacing the Airstrass either. Integration by parts results in an infinite integral series. $\End Group$ 4 The solution to the integral part of sin^2(x) requires you to recall the principles of trigonometry and calculus. Do not conclude that because the sin integral (x) equals cos(x), the integral of sin^2(x) should be equal to -cos^2(x); In fact, the answer contains no cochinus at all. You cannot directly combine sin^2(x). Use trigonometric identities and calculus substitution rules to resolve the issue. Use a half-angle formula, sin^2(x) = 1/2*(1 - cos(2x)) and switch integral so that it becomes 1/2 times inseparable from (1 - cos(2x)) dx. Set u = 2x and du = 2dx to make u substitution on the integral. Since dx = du/2, the result is 1/4 times the integral of (1 - cos(u)) du. Combine the equation. Since the integral of 1du is u, and the integral of cos(u) du is sin(u), the result is 1/4*(u - sin (u)) + c. Replace u back into equation to get 1/4*(2x - sin(2x)) + c. Simplify to get x/2 - (sin(x))/4 + c. Tips for a definite integral, turn off the constant in the reply, and edit the answer along the interval specified in the problem. If the interval is 0 to 1, for example, evaluate [1/2 - Sin(1)/4] - [0/2 - Sin(0)/4)].

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