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The variable appearing in a definite integral is known as a dummy variable, which means $\int_a^b f(x)dx$ is the same thing as $\int_a^b f(t)dt$. Now this result comes from the fact that we are simply substituting $x=t$. If we substitute anything that is not equal to $x$, everything changes. For example,

Consider the following identity-

$$\int_a^bf(x) dx=\int_a^bf(a+b-x)dx $$

The following proof has been provided in my textbook-

Substitute $x=a+b-t$. Then,

$$\int_a^bf(x) dx=-\int_b^af(a+b-t)dt$$ The limits have changed because we have substituted a variable in place of $x$ which is strictly not equal to $x$

$$=\int_a^bf(a+b-t)dt$$

$$=\int_a^bf(a+b-x)dx$$

This is the step where I have the problem. The concept of dummy variable comes from the fact that we can substitute variable of integration=any variable. So in this case we are substituting $t=x$. But previously we had claimed that $x=a+b-t$. Isn't this a contradiction and thus not a proper proof?

Disintegrating By Parts
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Anurag Saha
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1 Answers1

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Substitution of a number or an expression for another number or expression of equal value is not the same as renaming variables.

What I mean by substitution is the following. If you have an equation, like $x^2 + y^2 = c$ and another equation giving the value of one of these variables, like $x = 2 + \pi$, then we can substitute $x$ for $2+\pi$ in $x^2 + y^2 = c$ to get $(2+\pi)^2 + y^2 = c$. This is known to work because it is known that the $x$ in $x^2 + y^2 = c$ equals $2+\pi$.

Variable renaming is not based on equality. It is more like a transformation. To illustrate what I mean by a transformation, suppose we started with the same equation $x^2 + y^2 = c$ and we wanted to make the transformation $x\to x+5$ where we replace every $x$ with $x+5$. Then after the transformation, we would have $(x+5)^2 + y^2 = c$. Clearly, $x\neq x+5$. Also, even if $x^2 + y^2 = c$ is true, it is not necessarily true that $(x+5)^2 + y^2 = c$. By this, you can see that transformations are not the same as substitutions, even if in some cases they look alike.

So in the last step, where you are trying to figure out how $\int_a^b f(a+b-t)dt=\int_a^b f(a+b-x)dx$, note that they are equal because you can perform a variable renaming operation (which is what I refer to above as a transformation), but that does not necessarily imply that $t=x$. Rather, you are simply performing the transformation $t\to x$.

Note that this variable renaming operation is in the lambda calculus and is called $\alpha$-conversion.

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