I am told that an expression like $$ \int_a^x f(x)dx $$ is not well formed, i.e. it should be $$ \int_a^xf(t)dt $$ or similar.
Why is it that the limits of integration can't depend on the variable of integration?
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4$\begingroup$ It's a bit like writing $\sum_{i = 1}^i f(i)$. What is it supposed to mean? $\endgroup$– Dylan MorelandCommented Feb 14, 2012 at 0:17
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4$\begingroup$ There is no real reason why it shouldn't (in the end the integration variable is a dummy variable) -- however, choosing the integration variable to be the same as one of the limits opens many doors for error... $\endgroup$– FabianCommented Feb 14, 2012 at 0:18
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2$\begingroup$ Try $\frac{d}{dx}$ on the first one. No confusion? $\endgroup$– Raymond ManzoniCommented Feb 14, 2012 at 0:19
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3$\begingroup$ You can do as you have in the first integral, but you shouldn't. The x's play entirely different roles when you use x to denote both a limit and a dummy variable $\endgroup$– The Chaz 2.0Commented Feb 14, 2012 at 0:20
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1$\begingroup$ Poor style, but in principle OK. $\endgroup$– André NicolasCommented Feb 14, 2012 at 0:29
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3 Answers
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In mathematics, it's generally regarded as a bad idea for the same symbol to have two different meanings in the same expression. In this case, the variable being integrated with respect to effectively disappears, and a new variable (really two new variables, the bounds of integration) takes over. To call them the same thing can make things confusing sometimes (although not always). This is more of a stylistic than a strictly logical concern, at least in one variable.
answered Feb 14, 2012 at 0:21
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It confuses a free variable and a bound variable. In effect you are saying "let $x$ range from $a$ to $x$ when taking the integral of $f(x)$".
It is also ambiguous. There is a risk some people might expect $\int_a^x f(x)dx = (x-a)f(x)$ in the same way as $\int_a^x f(x)dt = (x-a)f(x)$.
It is easier to show the problem as a sum. The sum of the first $n$ positive integers can be written $\sum_{i=1}^{i=n} i = \frac{n(n+1)}{2}$ but if you wrote it as $\sum_1^n n$, some people might expect the answer to be $n^2$. Meanwhile the following looks very strange $$1+2+3+\cdots+n+\cdots+(n-1)+n$$
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$\begingroup$ I have seen that abuse of notation used in physics. For example, in kinematics, $x(t) = \int_0^t v(t) d t$. It is clearer than using a non-standard symbol for time. $\endgroup$– mlainzCommented Jun 3, 2019 at 14:23
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your question is very valid and I see all the answers posted say a very comfortable no. I can give a physical situation where the limit is also a function of the variable in the integral (one which I am trying to solve myself). It goes like this:
The solar radiation intensity ($W/m^2$) at a point on the surface of the earth depends on the angle (theta) between the incident ray and the normal (radius vector) at the point. The exercise is to find the total radiant energy ($J/m^2$) over the period of a year. Now, theta as a function of time varies continuously and the limit of integration would cover the longitudinal sweep from sunrise to sunset. Bear in mind that this longitudinal sweep of daylight (from sunrise to sunset) varies with latitude of the point and also the time of year; thus the daily limit of integration is also a function of time (and this has to be summed up over a year).
I hope the problem statement is clear and I have worked out the angle relationships between theta, latitude, tilt of earth axis and angle swept by the earth in its orbit around the sun (taking the northern hemisphere summer solstice as the starting point).
It is a very practical problem and I'm sure that a solution to this exists that may have been discovered or is yet to be discovered.
