Calcolatore di integrali definiti e impropri

Il calcolatore cercherà di valutare l'integrale definito (cioè con estremi), inclusi quelli impropri, mostrando i passaggi.

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Integrate with respect to:

Enter a lower limit:

If you need `-oo`, type -inf.

Enter an upper limit:

If you need `oo`, type inf.

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Solution

Your input: calculate $$$\int_{f m n^{2} \nu s t y}^{\infty}\left( \sin{\left(x^{2} \right)} \right)dx$$$

First, calculate the corresponding indefinite integral: $$$\int{\sin{\left(x^{2} \right)} d x}=\frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} x}{\sqrt{\pi}}\right)}{2}$$$ (for steps, see indefinite integral calculator)

Since there is infinity in the upper bound, this is improper integral of type 1.

To evaluate an integral over an interval, we use the Fundamental Theorem of Calculus. However, we need to use limit if an endpoint of the interval is special (infinite).

$$$\int_{f m n^{2} \nu s t y}^{\infty}\left( \sin{\left(x^{2} \right)} \right)dx=\lim_{x \to \infty}\left(\frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} x}{\sqrt{\pi}}\right)}{2}\right)-\left(\frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} x}{\sqrt{\pi}}\right)}{2}\right)|_{\left(x=f m n^{2} \nu s t y\right)}=- \frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} f m n^{2} \nu s t y}{\sqrt{\pi}}\right)}{2} + \frac{\sqrt{2} \sqrt{\pi}}{4}$$$

Answer: $$$\int_{f m n^{2} \nu s t y}^{\infty}\left( \sin{\left(x^{2} \right)} \right)dx=- \frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} f m n^{2} \nu s t y}{\sqrt{\pi}}\right)}{2} + \frac{\sqrt{2} \sqrt{\pi}}{4}$$$

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Solution