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Finding value of $\displaystyle \int^{\infty}_{0}\sin(t)dt$

What i have tried yet

As we know that period of $\sin(x)$ is $2\pi$

So we can split the intehral as

$\displaystyle \lim_{n\rightarrow \infty}\bigg[\int^{2\pi}_{0}\sin(t)dt+\int^{4\pi}_{2\pi}\sin(t)dt+\int^{6\pi}_{4\pi}\sin(t)dt+\cdots\cdots +\int^{2n\pi}_{(2n-1)\pi}\sin(t)dt +\int^{(2n+2)\pi}_{2n\pi}\sin(t)dt\bigg)\bigg]=0$

Because $\displaystyle \int^{2\pi}_{0}\sin(t)dt=0$

What i have mention is right, if not then please explain me , thanks

jacky
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3 Answers3

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The integral is undefined (aka. "divergent").

To see this, you need to consult the definition of a definite integral with upper bound $\infty$:

$$\int_{0}^{\infty} f(x)\ dx := \lim_{x_\mathrm{max} \rightarrow \infty} \int_{0}^{x_\mathrm{max}} f(x)\ dx$$

This limit is a limit over the real numbers, and hence your problem is equivalent to asking what the value of

$$\lim_{x_\mathrm{max} \rightarrow \infty} \int_{0}^{x_\mathrm{max}} \sin(x)\ dx$$

is. But since the right-hand integral is just $1 - \cos(x_\mathrm{max})$, then that means we are asking for

$$\lim_{x_\mathrm{max} \rightarrow \infty} [1 - \cos(x_\mathrm{max})]$$

And this limit does not exist since the trig always oscillates up and down periodically; hence also, the integral has no value.

The_Sympathizer
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  • Thanks so much The_Sympathizer. Would u please explain me where i am wrong. – jacky Apr 05 '20 at 03:21
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    @jacky: When you try to split up the integral, you are, in effect, presuming that a value exists to begin with so as to do that splitting. Alternatively, consider what would happen if you split it up in a slightly different way - e.g. intervals of width $2\pi + 1$. – The_Sympathizer Apr 05 '20 at 03:23
  • @Eevee Trainer: When you apply the FTC, you have to use the antiderivative at both bounds. So it should be $[-\cos(x_\mathrm{max})] - [-\cos(0)]$ - note $\cos(0)$, not $\sin(0)$. – The_Sympathizer Apr 05 '20 at 03:25
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The_Sympathizer explained what's wrong with your approach in a comment. Here's an example of how your reasoning could be abused: \begin{align*} \int_0^\infty \sin(t) \, \mathrm{d}t &= \int_0^\pi \sin(t) \, \mathrm{d}t + \left(\int_\pi^{3\pi} \sin(t) \, \mathrm{d}t + \int_{3\pi}^{5\pi} \sin(t) \, \mathrm{d}t + \ldots\right) \\ &= 2 + \left(0 + 0 + \ldots\right) = 2. \end{align*} How can the same integral equal two different values? If the improper integral exists, we would get the same number, no matter how we subdivided the integral. Thus, the combination of our two computations would prove that the improper integral is divergent, by contradiction.

user764828
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What you have shown correctly is that

$$\lim_{n\to \infty}\int_0^{2\pi n}\sin x \; dx = 0$$

The logical fallacy consists in your conclusion: $$\color{red}{\not \Rightarrow \lim_{b\to \infty}\int_0^{b}\sin x \; dx \text{ exists and is equal to }0 \text{ (Wrong!)}}$$

trancelocation
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