# $u$-substitution always evaluates to $0$

Consider the integral

$$\int_a^b f(x) \,dx$$

now make the $u$-substitution $u \mapsto c + (x-a)(x-b)$. The resulting integral is

$$\int_c^c h(u) \,du$$

where $h(u)$ is the integrand $f$ after the substitution, however, regardless of $f$ the integral $\int_c^c du = 0$. Looking at the definition Wikipedia provides I believe the substitution meets every condition. It's differentiable and has a integrable derivative, because it's a polynomial. So what's wrong with this substitution such that it always results in $0$?

• It needs to be bijective Jun 16, 2017 at 16:10
• @BrevanEllefsen : I think the OP want to use the formula $\int_{\phi(a)}^{\phi(b)}f(t)\,dt=\int_a^bf(\phi(t))\phi'(t)\,dt$, which is right even if $\phi$ is only $\mathcal{C}^1$ ? Jun 16, 2017 at 16:14
• @Baloown Yes, that seems correct, as long as $\phi$ is integrable. The reason for the issue here is clearly the lack of a bijection, which you explain well in your answer :) Jun 16, 2017 at 16:17

The formula $$\int_{g(a)}^{g(b)} f(x) \, dx=\int_{a} ^{b} f(g(t)) g'(t) \, dt$$ holds without $g$ being bijective. But as you can see the substitution has to be of the form $x=g(t)$ and you are trying to make a substitution using $t=h(x)$ and this would require that $h$ is invertible to put it in the form $x=h^{-1}(t)$ so that the above Rule for change of variables applies. Your substitution does not have this property and hence the conclusion obtained is wrong.

It is better to understand the theorems completely before applying them. And more often than not most people remember only the conclusions of the theorems rather than their hypotheses. By habit one should avoid this and understand how the hypotheses lead to conclusions and what happens when one of the hypotheses is not satisfied.

• Fantastic observation. Jun 17, 2017 at 4:55
• @BrevanEllefsen : I have seen many such questions here and most answers deal with bijective nature of substitution which is not really necessary (at least not in the way the change of variables is stated as a theorem). People don't often use the theorem as stated. Jun 17, 2017 at 5:15
• But Paramanand, the OP did in fact make an explicit, not an implicit, substitution: they used the subsitution $u=g(x)$ to transform to transform their integral from the $x$-world to the $u$-world. I've just posted an answer explaining their mistake. Feb 8 at 16:10
• @ryang: the change of variables in my answer is of the form $x=g(t)$ and not $t=g(x)$ and we don't need $g$ being injective. Feb 9 at 5:19
• @ParamanandSingh The issue is this: your above answer mistakenly accuses the OP's of making a non-invertible implicit substitution (their substitution $u=g(x),$ where $x$ is their starting variable, is in fact non-invertible but not an implicit substitution), so falsely concludes that the OP's work is invalid. Feb 11 at 10:44

The formula of change of variables is, for $\phi\in\mathcal{C}^1$, $$\int_{\phi(a)}^{\phi(b)}f(t)\,dt=\int_a^bf(\phi(t))\phi'(t)\,dt,$$ so to get what you write, you should have that $f$ is written as a function of the form $$\tilde{f}\circ \text{“your function''}\times\text{“the derivative of your function''},$$ or to inverse "your function", which is of course not possible since it is not bijective.

EDIT : More precisely, if you want to change both the integrand and the boundaries, you have to find $\tilde{a}\in\phi^{-1}(a)$ and $\tilde{b}\in\phi^{-1}(b)$ to finally write $$\int_a^bf(t)\,dt=\int_\tilde{a}^\tilde{b}f(\phi(t))\phi'(t)\,dt,$$ and if $a\neq b$ then $\tilde{a}\neq\tilde{b}$ (else $a=\phi(\tilde{a})=\phi(\tilde{b})=b$), and you can't conclude a priori that the right integral is $0$ using the same boundaries argument.

• what needs to be bijective? certainly not $\phi(x)$, because to solve $\int x \cos(x^2 + 1) \,dx$, we would use $\phi(x) = x^2 +1$ as a substitution which is not bijective. Jun 16, 2017 at 16:17
• @Dando18 within certain intervals it is bijective. There are a lot of potential caveats here in that anti-derivatives can actually vary on different intervals where substitutions work differently Jun 16, 2017 at 16:18
• is $c+(x-a)(x-b)$ not bijective on certain intervals? Jun 16, 2017 at 16:19
• @Dando18: I think that you are trying to use the formula $$\int_a^bf(t)\,dt=\int_{\phi^{-1}(a)}^{\phi^{-1}(b)}f(\phi(t))\phi'(t)\,dt,$$ which allows you to change both integrand and boundaries, but as you see explicitely in the formula in this case you have to consider a $\phi$ which is invertible. Jun 16, 2017 at 16:20
• @Dando18 What is required is that the substitution be bijective on the interval you are integrating on. Your work above shows that your function isn't bijective, so we are done. Well, sorta. There are very specific cases where we can do this when our function is not bijective, but those are extremely tricky and nuanced and best handled in my opinion by splitting the domain into bijective sections Jun 16, 2017 at 16:21

$$\int_a^b f(x) \,\mathrm dx$$ now make the $$u$$-substitution $$u \mapsto c + (x-a)(x-b).$$

Typo: judging from your post, your substitution is actually \begin{align}u=g(x)&=c + (x-a)(x-b)\tag# \\ g'(x)&=2x-a-b\\g(a)&=c=g(b).\end{align}

The resulting integral is $$\int_c^c h(u) \,\mathrm du =0.$$

what's wrong with this substitution such that, regardless of the starting integrand $$f,$$ it results in $$0\ ?$$

It is indeed correct that $$\int_a^b f(x) \,dx\overset{u = c + (x-a)(x-b)}{=}0.\tag✓$$ However, this is not “regardless of $$f,$$” but whenever the substitution is even possible, which is precisely when $$f$$ has rotational symmetry about $$\left(\frac{a+b}2,0\right)$$ and $$h$$ is integrable, and has an antiderivative, on $$g[a,b].$$ (The stipulation on $$h$$ means that the red bit in $$\int_c^c \color{red}{\text{[this integrand doesn't matter]}} \,dx=0$$ is false.) So, if $$f$$ lacks the stipulated symmetry property, then there is no basis for, and no way to, make the given substitution $$(\#).$$

Proof

Let's begin with this strong version of the integration-by-substitution theorem, which does not generally require injectivity:

• If $$g'$$ is integrable on $$[a,b]$$ and $$h$$ is integrable, and has an antiderivative, on $$g[a,b],$$ then $$\int_a^bh\big(\color{violet}{g(x)}\big)\,\color{cyan}{g'(x)}\,\mathrm{d}x=\int_{g(a)}^{g(b)}h(u)\,\mathrm{d}u.$$

This corresponds exactly to your above work, with $$f(x)=h\Big(\color{violet}{c + (x-a)(x-b)}\Big)\,\color{cyan}{(2x-a-b)}.$$ As $$f$$ has not been concretely given, function $$h$$ is necessarily abstract (though we note that the choice of $$h$$ is also to satisfy the theorem's conditions; this immediately happens if $$h$$ is continuous on $$g[a,b].)$$ Now let's shift $$f$$ by $$\displaystyle\frac{a+b}2$$ units leftwards: $$f_N(x)=f\left(x+\frac{a+b}2\right)\\ =h\Bigg(c + \left(\left(x+\frac{a+b}2\right)-a\right)\left(\left(x+\frac{a+b}2\right)-b\right)\Bigg)\:\:\left(2\left(x+\frac{a+b}2\right)-a-b\right)\\ =h\left(c+x^2-\frac{(a-b)^2}4\right)\;\;2x.$$ Since $$f_N(-x)=-f_N(x),$$ the new function is odd. Thus, $$f$$ has rotational symmetry about the point $$\left(\frac{a+b}2,0\right).$$

In the youtube video Weird Integral Trick, @blackpenredpen demonstrates that $$u=g(x)=x(x-a-b)$$ is a wonderful substitution, because it amazingly and universally zeroes out every integral: $$\int_a^b f(x)\,\mathrm dx\overset{u=x(x-a-b)}{=}\int_{-ab}^{-ab}\text{[this integrand doesn't matter]}\,\mathrm du=0.$$

The apparent anomaly is currently explained in neither the video nor its comments, so, for the record, it has the same explanation as above: this substitution is possible precisely when both these conditions are satisfied: 1) $$f$$ on $$[a,b]$$ has rotational symmetry about $$\left(\frac{a+b}2,0\right),$$ 2) $$h$$ on $$g[a,b]$$ is integrable and has an antiderivative; if so, then the result $$0$$ is correct.

• Right now I can only say that the theorem in your answer is the same as in my answer. I don't understand why we are reaching to somewhat different conclusions. I will have a deeper look at both answers and comment more later. Voting on this will be delayed Feb 9 at 5:22
• If the original integral is in $x$ domain then the substitution must be of the form $x=g(u)$. If the substitution $u=g(x)$ is used we need $g$ injective. Feb 11 at 1:40
• @ParamanandSingh Now you're contradicting yourself: your posted answer correctly explains that (paraphrasing for clarity) a substitution must be invertible iff the new variable is an implicit function of the starting variable (e.g., $x=\sin u,$ where $x$ is the starting variable), $\quad$ but your above comment is now incorrectly claiming the opposite. Feb 11 at 10:42
• @ParamanandSingh The issue is this: your posted answer mistakenly accuses the OP's of making a non-invertible implicit substitution (their substitution $u=g(x),$ where $x$ is their starting variable, is in fact non-invertible but not an implicit substitution), so falsely concludes that the OP's work is invalid. Feb 11 at 10:43
• I finally understood your point here (+1). Sorry for the delay. The thing is that the substitution works fine not for all $f$ but for all $h$ and $f(x)$ should be of the form $f(x) =h(c+(x-a) (x-b))$. Finally looking at both answers (yours and mine) I conclude that the theorems used here are same and results obtained are therefore correct. Aug 18 at 0:27

Assume $a<b$.

If $u=c+(x-a)(x-b)$. Then

\begin{align} u&=x^2-(a+b)x+ab+c\\ &=\left(x-\frac{a+b}{2}\right)^2+ab-\left(\frac{a+b}{2}\right)^2+c\\ &=\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2+c\\ x&=\frac{a+b}{2}\pm\sqrt{\left(\frac{a-b}{2}\right)^2-c-u} \end{align}

When $\displaystyle x<\frac{a+b}{2}$, $\displaystyle x=\frac{a+b}{2}-\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}$ and

$$\frac{du}{dx}=2x-(a+b)=-\sqrt{(a-b)^2-4c-4u}$$

When $\displaystyle x\ge\frac{a+b}{2}$, $\displaystyle x=\frac{a+b}{2}+\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}$ and

$$\frac{du}{dx}=2x-(a+b)=\sqrt{(a-b)^2-4c-4u}$$

When $\displaystyle x=\frac{a+b}{2}$, $\displaystyle u=c-\left(\frac{a-b}{2}\right)^2$

Therefore,

\begin{align} \int_a^bf(x)dx&=\int_a^{\frac{a+b}{2}}f(x)dx+\int_{\frac{a+b}{2}}^bf(x)dx\\ &=\int_0^{c-\left(\frac{a-b}{2}\right)^2}\left[\frac{f\left(\frac{a+b}{2}-\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}\right)}{-\sqrt{(a-b)^2-4c-4u}}\right]du\\ &\qquad +\int_{c-\left(\frac{a-b}{2}\right)^2}^0\left[\frac{f\left(\frac{a+b}{2}+\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}\right)}{\sqrt{(a-b)^2-4c-4u}}\right]du \end{align}

It looks quite complicated. But my point is that the definite integral should be broken into two parts with different integrands.

\begin{align} u & = c + (x-a)(x-b) \\[10pt] &= c + \left( x - \frac{a+b} 2 \right)^2 + \left( \frac{a-b} 2 \right)^2 & & \text{(completing the square)} \\[12pt] & \text{When } x = a \text{ or } x=b \text{ then } u = c. \\ & \text{When } x = \frac{a+b}2 \text{ then } u = c+ \left( \frac{a+b} 2 \right)^2. \end{align} $$\int_a^b f(x)\,dx = \int_a^{(a+b)/2} f(x) \,dx + \int_{(a+b)/2}^c f(x)\, dx\quad\longleftarrow \text{two terms}$$ $$du = 2\left( x - \frac{a+b} 2 \right) \, dx = \Big( \text{some function of } u \Big)\, dx$$ This "some function" is one thing in one of the "two terms" above and is another thing in the other one of the "two terms" above. The reason is that when you find $x-\dfrac{a+b}2$ as a function of $u,$ you get the usual “plus-or-minus” $(\text{“}\pm\text{''})$ issue that you see when solving quadratic equations. In the first of the "two terms" above, you need the solution with the minus sign, and in the second the one with the plus sign.

Your error is thinking that "the integrand $f$ after substitution" is well defined. (actually, the integrand is $f(x) \mathrm{d} x$, but that distinction is not relevant to the topic)

It is indeed true that if $f$ is a function with the property that there exists a function $h$ such that

$$f(x) \mathrm{d} x = h(u) \mathrm{d}u$$

(where $u$ and $x$ are related by $u = c + (x-a)(x-b)$), then you do have

$$\int_a^b f(x) \mathrm{d}x = \int_c^c h(u) \mathrm{d}u = 0$$

However, you are not guaranteed such an $h$ exists. Instead, the generally $2$ to $1$ nature of the relationship between $x$ and $u$ means that on any path where $x$ varies from $a$ to $b$, you will get one function $h_0$ on part of the path and another function $h_1$ on the rest of the path. Then, the best you can say is that

$$\int_a^b f(x) \, \mathrm{d}x = \int_c^d h_0(u) \mathrm{d}u + \int_d^c h_1(u) \mathrm{d}u$$

• interesting, would it not be possible to express $h(u)$ in one function using special functions or $\mathrm{sgn}(u)$? Or would that affect our ability to integrate $h(u)$? Jun 17, 2017 at 13:12