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定积分与广义积分计算器
逐步计算定积分和广义积分
该计算器将尝试计算定积分（即带上下限的积分），包括广义积分，并显示求解步骤。
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: $$$\int_{f m n^{2} \nu s t y}^{\infty}\left( \sin{\left(x^{2} \right)} \right)dx=- \frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} f m n^{2} \nu s t y}{\sqrt{\pi}}\right)}{2} + \frac{\sqrt{2} \sqrt{\pi}}{4}$$$