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My book say that integration of $x^{1/2} \sin x$ is not possible, why is it so? Which functions do not have an anti derivative?
Does it mean that they do not have any area under the curve?
But that's not true since the graph says different.enter image description here


(Source: https://www.desmos.com/calculator)

When is a function not integrable?

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    $\begingroup$ The integral is not expressible in terms of elementary functions. It can be integrated and the integral is perfectly defined, but the result may have to be expressed by some other specially defined function. $\endgroup$ – ZTransformer Jan 2 '16 at 23:19
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    $\begingroup$ Is your question about integration by parts, about integrable functions or about closed-form antiderivatives ?? $\endgroup$ – Yves Daoust Jan 2 '16 at 23:23
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    $\begingroup$ What is the meaning of the title then ? And what should we understand by "My book say it is not possible" ? Please improve your post. $\endgroup$ – Yves Daoust Jan 2 '16 at 23:25
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    $\begingroup$ This is simply not true, consider the function $f(x) = \int_0^x \sqrt{t} \sin t \,dt$. By FTC its first derivative is the given function $\sqrt{x} \sin x$. There are pathological functions that are not integrable in one sense or another, but, e.g., all continuous functions admit antiderivatives. $\endgroup$ – Travis Jan 2 '16 at 23:25
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    $\begingroup$ If that is all and exactly what the book says,it is BS. Every continuous real function $f$ has an anti-derivative. That is part of the Fundamental Theorem of Calculus :$(d/dt)\int _0^t f(x) dx=f(t).$ Some ordinary-looking continuous functions have anti-derivatives that cannot be expressed in simple formulas using more basic functions. $\int \sqrt x \sin x dx $ looks like one of these. $\endgroup$ – DanielWainfleet Jan 3 '16 at 4:47
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$$\int x^{1/2}\sin(x)dx=\int x^{1/2}\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!} dx=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+5/2}}{(2k+\frac52)(2k+1)!}.$$

The series converges for all $x$.


By substitution then by parts, you have

$$\int x^{1/2}\sin(x)dx=2\int t^2\sin(t^2)dt=-t\cos(t^2)+\int \cos(t^2)dt,$$ where the last integral is known as the Fresnel cosine function.


To get a numerical value of the definite integral, Simpson's method will perform better.

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There is an important distinction between integrable and expressible in terms of elementary functions. Integrable means that the integral exists. If we take one of the endpoints to be variable then we get the anti derivative. All continuous functions such as the one you mention are integrable. The question of whether this antiderivative can be written as a combination of the usual rational, algebraic, exponential, logarithmic, and trigonometric functions is different. The antiderivitive of your function exists, but cannot be so expressed.

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  • $\begingroup$ Can you help me with integrating this function? $\endgroup$ – Quark Jan 2 '16 at 23:24
  • $\begingroup$ No and neither can anyone else. $\endgroup$ – Rene Schipperus Jan 2 '16 at 23:27
  • $\begingroup$ How do i find the area under this curve between x=2 to x=10 $\endgroup$ – Quark Jan 2 '16 at 23:30
  • $\begingroup$ There are numerical methods and other approximations. I am not the right person to ask, you could enter another MSE question about the best way to find a numerical answer. $\endgroup$ – Rene Schipperus Jan 2 '16 at 23:33

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