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I tried to find the integral $\int f(x)dx$ of $f(x)=\sqrt{1+\dfrac{1}{x^4}}$. I couldn't solve this manually and used https://www.integral-calculator.com/ for finding the integral. Surprisingly, a message was displayed.

Antiderivative or integral could not be found. Note that many functions don't have an elementary antiderivative.

Now, I was confused as integral $\int f(x)dx$ represented the area under the curve $y=f(x)$. Now, the curve $y=\sqrt{1+\dfrac{1}{x^4}}$ has some area under it. So, why doesn't it have an integral?

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    $\begingroup$ The map has an antiderivative, but the antiderivative can't be written with elementary maps. $\endgroup$ Jun 15, 2022 at 14:12
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    $\begingroup$ "There are ways". It's not always necessary that the function has an antiderivative as an elementary function, but the area under the curve exists. $\endgroup$
    – user983440
    Jun 15, 2022 at 14:18
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    $\begingroup$ For instance, exploit the geometry of the shape whose area you are trying to compute. A classic example is $\int_{-\infty}^\infty e^{-x^2} \, d x = \sqrt{\pi}$. The integrand does not have an elementary antiderivative, but all the same the (improper!) definite integral is perfectly computable by some cute tricks. (See en.wikipedia.org/wiki/Gaussian_integral). $\endgroup$ Jun 15, 2022 at 14:22
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    $\begingroup$ See the anti-derivative in terms of non-elementary function here if you are interested. $\endgroup$
    – user983440
    Jun 15, 2022 at 14:24
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    $\begingroup$ Lost in this discussion is what it means for a function to be non-elementary. Before learning about logs, one might assume that every function can be expressed as some finite composition of polynomials, exponentials, and kth-roots. Only when they try to invert e^x would they realize that the inverse-exponentials (logarithms) define a brand new class of functions! But the inverse-exponential was still a perfectly valid function even before assigning it the name "log". The same thing is going on here: we don't have a cool name for the fn defined by your integral, but the fn certainly exists! $\endgroup$
    – Yonatan N
    Jun 15, 2022 at 22:24

4 Answers 4

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The indefinite integral $\displaystyle \int f(x) \, d x$ represents the collection of all functions whose derivatives with respect to $x$ equals $f(x)$---they're antiderivatives. This is in contrast with the definite integral $\displaystyle \int_a^b f(x) \, d x$ which represents the area under the curve $y = f(x)$ from $x = a$ to $x = b$.

These objects are related by the Fundamental theorem of calculus which says (in one of its many forms) that the area in the second expression can be computed by evaluating any one of the antiderivatives at the endpoints and subtracting.

Note that it does not say that using antiderivatives is the only way to compute the area under the curve, it's just a very convenient way in many situations. There are many situations in which it is impossible to write down a nice expression for the indefinite integral (e.g. in terms of elementary functions) but nevertheless possible to compute the area just fine. One way of doing this is to exploit the geometry of the shape whose area we are after.

A classic example (mentioned in comments) is the Gaussian integral $$ \int_{-\infty}^\infty e^{-x^2} \, d x = \sqrt{\pi}. $$ There are also examples where it is perfectly possible to find a nice antiderivative, but it may be difficult, or at least more difficult than some other approach. For instance, $$ \int_{-1}^1 \sqrt{1 - x^2} \, d x = \frac{\pi}{2}. $$ We could compute this by finding an antiderivative (which is doable but tricky if we're not familiar with trigonometric substitutions), but it is (probably) easier to notice that the shape we are after is really a semicircle. (You may argue that this is circular, depending on how you derived the formula for the area of a circle in the first place.)

As to why some functions don't have elementary antiderivatives: this is a fairly deep question to do with an area of mathematics called differential algebra. A reasonably approachable exposition I enjoy is Brian Conrad's Impossibility theorems for elementary integration.

Finally there are powerful methods to approximate definite integrals---to estimate the area under the curve $y = f(x)$---without having a clue whatsoever how one might find an antiderivative. This is an entire area of mathematics in its own right called numerical integration. Depending on what one is doing, this may be all one needs.

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    $\begingroup$ +1 for good answer, comment for the chuckle I got at the idea of using the area of a semi-circle being circular reasoning $\endgroup$
    – Alan
    Jun 15, 2022 at 15:18
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One thing which has not been mentioned explicitly in the comments or the other answer is the following fact:

Your function $f(x)=\sqrt{1+\dfrac{1}{x^4}}$ does have an indefinite integral, i.e. an antiderivative, on any interval where it is defined.

For example it has an antiderivative on the interval $0 < x < +\infty$ (we have to avoid $x=0$ because it is not in the domain of $f(x)$).

That fact is true by application of the first fundamental theorem of calculus. For your particular example, that theorem tells you that the formula $$F(x) = \int_1^x \sqrt{1+\dfrac{1}{t^4}} \, dt $$ defines an antiderivative of your function $f(x)$ for all values $0 < x < \infty$.

Now, you might not particularly like that formula for $F(x)$, because it's hard to evaluate.

What you can always do is to use that formula for $F(x)$ to get a numerical estimate of its output value for any input value $x \in (0,\infty)$, by using the trapezoidal rule for example, or any other numerical integration method of your choice.

What you can sometimes do for other functions $f(x)$ is to use methods of integration to write down an explicit elementary formula for $F(x)$.

But for this particular function $f(x) = \sqrt{1+\dfrac{1}{x^4}}$ you cannot find an elementary formula for $F(x)$. See the other answer for a fuller explanation of this point.

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the integral

$$ \int\sqrt{1+\frac{1}{x^4}}dx $$

can be rewritten as $$ \int x^{-2}(1+x^4)^{\frac{1}{2}}dx, $$

which is an example of integrating the differential binom.

In this case, the integral (antiderivative) can be expressed in elementary functions only if one of the following holds:

  • $\frac{-2 + 1}{4} + \frac{1}{2} \in \mathbb{Z} \Leftrightarrow \frac{1}{4} \in \mathbb{Z}$ (FALSE)
  • $\frac{-2 + 1}{4} \in \mathbb{Z} \Leftrightarrow -\frac{1}{4} \in \mathbb{Z}$ (FALSE)
  • $\frac{1}{2} \in \mathbb{Z}$ (FALSE)

Non of the above is TRUE, so the integral cannot be expressed in elementary functions.

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    $\begingroup$ Why those formulations exactly? $\endgroup$
    – No Name
    Jun 15, 2022 at 22:41
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    $\begingroup$ Your answer does not really make sense. For example, what is the significance of the expression $\frac{-2 + 1}{4} + \frac{1}{2}$? $\endgroup$
    – Lee Mosher
    Jun 16, 2022 at 13:01
  • $\begingroup$ If you follow the link I put in the answer, you will see why just these scenarios. Also, here, you can read what to do if a particular case is true. $\endgroup$
    – Eugene
    Jun 16, 2022 at 17:02
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To answer the question in the title (completely separate from the question of elementary-function antiderivatives):

Does every function have an integral?

No, there are functions that do not have any defined integral at all. But the exact category of such functions depends on which definition of integral you're using: Riemann, Lebesgue, or less commonly, something else.

For example, the characteristic function of the rational numbers (function with value 1 at a rational number, 0 elsewhere) is not Riemann integrable, but it is Lebesgue integrable. On the other hand, the sinc function taken over the entire real number line is not Lebesgue integrable.

If you're interested in pursuing that more, consider reading the Wikipedia article on the Lebesgue integral, particularly sections on limitations of the various integral definitions.

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