# How does this integral evaluate to $0=1$?

While looking on forums for challenging high school math questions, I came across the following integral:

(excuse me if I forget to write dt or dx, the work should still be correct)

$$I=\int\frac{2x^{2}+1}{x^{3}+x\ln\left(x\right)} \,\Bbb dx.$$

First I used the substitution $$\ln(x)=t$$ leading to the following: $$x=e^t, \qquad \frac{\Bbb dx}{\Bbb dt}=e^t.$$

\begin{align} I&=\int \frac{2e^{2t}+1}{e^{3t}+t\cdot e^{t}}\,\Bbb dx\\ &=\int\frac{2e^{2t}+1}{e^{2t}+t}\,\Bbb dt \tag{Eq. 1}\\ &=\int\frac{f'(t)}{f(t)}\,\Bbb dt\\ &=\ln(e^{2t}+t)+c. \tag{Integral exists} \end{align}

This proves the existence of the integral. Continuing from (Eq. 1) and integrating by parts yields the following:

$$(\text{Eq. 1}) \implies I=\int\frac{2e^{2t}+1}{e^{2t}+t}\,\Bbb dt =\int uv'=uv-\int vdu.$$

$$u=(e^{2t}+t)^{-1}$$, $$u'=-(e^{2t}+t)^{-2}\cdot(2e^{2t}+1)$$.

$$v'=2e^{2t}+1$$, $$v=e^{2t}+t$$.

\begin{align} I&=(e^{2t}+t)\cdot(e^{2t}+t)^{-1}-\int -(e^{2t}+t)^{-2}\cdot(2e^{2t}+1)\cdot(e^{2t}+t)\\ &=\frac{(e^{2t}+t)}{(e^{2t}+t)}+\int(e^{2t}+t)^{-1}\cdot(2e^{2t}+1)\\ &=1+\int\frac{2e^{2t}+1}{e^{2t}+t}, \end{align} and $$\int(2e^{2t}+1)/(e^{2t}+t)\,\Bbb dt$$ is just equal to $$I$$, so:

\begin{align} I &= 1+I, \\ 0 &= 1. \end{align}

As for any argument relating to $$I$$ diverging to infinity, thus making the statement true, I could just make the integral definite in $$(a,b)$$ for any values of a and b that are both defined and result in a finite value for the integral

Similarly, if anyone thinks it’s because of the constant, assume:

$$I=f(t) +c,$$

then

$$f(t) +c =1+f(t) +c,\\0=1$$

And even if you could evaluate them separately for two different constants $$C$$ and $$D$$ such that $$C=D+1$$ and they cancel out to $$0=1$$, I could always just do the opposite and assign initial conditions to evaluate it such that it again leads to a contradiction. Or I could apply limits so that the constants cancel out anyway.

I don’t think either of these arguments are the real solution. I couldn’t find any restrictions on using integration by parts, or any mistake in my work. Still stuck on this.

• The $+C$ in each does not need to be the same $C$. Sep 19, 2023 at 2:06
• At a glance you seem to be ignoring the difference between a definite integral and an indefinite integral. When you write $I = \int_{ }^{ }\frac{2e^{2t}+1}{e^{3t}+t\cdot e^{t}}dt$, you are most charitably describing that $I$ is a function (argument omitted) whose derivative is the expression given as an integrand. So it is an error to conclude that $I$ is a particular value like $0$ or $1$. Sep 19, 2023 at 2:09
• "I could just make the integral definite" - great idea. Try it. Sep 19, 2023 at 2:12
• To emphasize or rephrase again... the indefinite integral of a function is a class of functions all of whom are the same except for the constant added. The class of functions if you were to solve the integral the one way happens to be the same class of functions if you were to solve it the other way, even if the representation varies slightly. There is no contradiction here. Sep 19, 2023 at 2:18
• @David oh is it because making it definite causes the 1 to become (1-1) resulting in 0=0 and no contradiction? Sep 19, 2023 at 2:57

You are misunderstanding the equation $$\int udv= uv-\int vdu$$.

This is an equality between indefinite integrals. As I said in the answer to the duplicate question I linked

In the world of computing anti-derivatives, "=" means "differ by a constant"...

And since this is causing so much confusion I am going to introduce some notation. Integration by parts is

$$\int udv =_{ii}uv-\int vdu$$ where the $$=_{ii}$$ means "equality as anti-derivatives/indefinite integrals".

If you go check the wikipedia page on integration by parts it says: "This is to be understood as an equality of functions with an unspecified constant added to each side. Taking the difference of each side between two values ... gives the definite integral version", which is saying same thing, although they don't bother to annotate the equals sign.

So now, we can clarify that "$$0 = 1$$" really should be written as "$$0 =_{ii} 1$$", and it means that when you consider them as functions they differ by a constant, which is true. If that's not clear, please feel free to add a comment about it.

Going back to your original computation, we see that what you figured out is (for your function $$r(t)$$)

$$\int r(t) dt =_{ii} 1 + \int r(t) dt \,\,\,\,\,\,\text{ (1)}$$ and then you are doing manipulations that you believe will turn the $$=_{ii}$$ into a $$=$$, but they don't, not validly. (Note that you still can subtract the integral from both sides, and that gives you "$$0=_{ii}1$$", which, as noted, is true.)

The things you can do to turn it into a "$$=$$" equation are:

• take the difference between the values of both sides at $$x=a$$ and $$x=b$$, i.e. do definite integrals. As you say, there's no contradiction in this case,

or

• introduce the arbitrary constants. So, suppose you have some $$R(x)$$ such that $$R'(x) = r(x)$$. Then you can turn equation (1) into $$R(x) +C_1 = 1+R(x) + C_2$$ You could then, if you wanted to, impose some initial condition identically on both sides, and figure out the relationship between $$C_1$$ and $$C_2$$. But notice that the initial condition must be applied to the entirety of each side - after all, those are the functions that are equal.

In reading your arguments, I wonder if this way of thinking about it might help: it seems like you think the arbitrary constant lies in the $$\int$$ symbol, but really it is in the $$=_{ii}$$ symbol, and you need to make the arbitrary constant explicit not when you remove the $$\int$$ sign, but when you convert the $$=_{ii}$$ to an $$=$$.

• Thanks, this helped, My point was that the arbitrary constsnt wasn’t arbitrary at all due to the initial conditions defining I, but in that case that definition cannot be attributed to the integral on the right hand side, meaning it isnt I = 1 + I its I = 1+ (integral) where (integral) is a function w the same antiderivative but a diff conatsnt as it isnt bound by the initial conditions. also your answer helped me better understand the nature of indefinite integrals and their constants, thanks Sep 19, 2023 at 20:03
• Phew, glad we finally got somewhere! :-) I agree that thinking about exactly where we can apply initial conditions helped. Also, feel free to give this answer an up vote or a check mark. ;-) Sep 19, 2023 at 20:44