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# A doubt on integration by parts

My friend had the following doubt:

$$\int { \frac { dx }{ x\ln { x } } }$$

You might laugh at me and say this is a straightforward integration by substitution which gets us to

$$\ln { (\ln { x)+C } }$$

But what if we try integration by parts???

$$I=\int { \frac { dx }{ x\ln { x } } } =\frac { 1 }{ \ln { x } } \int { \frac { dx }{ x } } -\int { \left( -\frac { 1 }{ { (\ln { x } ) }^{ 2 } } \frac { 1 }{ x } \int { \frac { dx }{ x } } \right) dx } =\frac { 1 }{ \ln { x } } \ln { x } +\int { \frac { dx }{ x\ln { x } } } =1+I\quad \\ \Rightarrow I=I+1\Rightarrow 1=0$$

Please tell me where are we wrong?

Note by Krishna Jha
2 years, 10 months ago

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Yes the approach using integration by parts tells us that something is wrong, but you will be surprised that both answers are true!

As long as we are working with indefinite integrals, the key here is the constant of integration. You forgot to add a constant when you integrated by parts. we know that this constant is arbitrary, that is, you could use $$C+1$$ as constant of integration rather than $$C$$.

Therefore this is correct using either methods, and you can assure that by two ideas :

1) you can differentiate both answers, you will get the same initial function.

2) suppose you are dealing with definite integral: when you use the fundamental theorem of calculus, the excess "$$1$$" will be cancelled. · 2 years, 10 months ago

This is one of my favorite "fallacies" in calculus.

In a similar vein: How to show that $$\sin \theta = \cos \theta = 0$$.

$\int \sin \theta \cos \theta \, d \theta = \int \sin \theta \, d \sin \theta = \frac{1}{2} \sin ^2 \theta,$

$\int \sin \theta \cos \theta \, d \theta = - \int \cos \theta \, d \cos \theta = - \frac{1}{2} \cos^2 \theta$

Hence, $$\frac{1}{2} \sin^2 \theta = - \frac{1}{2} \cos ^2 \theta$$. But since squares are non-negative, they must both be 0.

Of course, as pointed out by Hasan, the correct conclusion should be:

$\frac{1}{2} \sin ^2 \theta + C_1 = - \frac{1}{2} \cos ^2 \theta + C_2 \Rightarrow \sin^2 \theta + \cos^2 \theta = C_2 - C_1,$

which is a well known fact. Staff · 2 years, 10 months ago

Yes, examples like these "fake" proofs are somehow interesting.

Another example is The integral $$\displaystyle \int \mathrm \tan x \sec^2 x \mathrm {d}x$$ . Using substitution $$u=\sec x$$ and $$u=\tan x$$ will yield two obviously different answers. · 2 years, 10 months ago

And wouldn't this imply that $$C_2-C_1=1$$? · 2 years, 10 months ago

How is integration of \sin \theta \cos \theta =sin^2(theta) ... if we assume \sin \theta = t, then we will get t^2/2 ...thus (sin theta)^2 / 2 ...and how can u write dsin(theta)...I understood the fallacies, but plz respond to my doubts · 2 years, 10 months ago

Thanks for pointing out the error. I've edited the comment accordingly. Sorry for the confusion.

As explained by @John Muradeli , I used the substitution of $$u = \sin \theta$$ and $$u = \cos \theta$$ respectively. It is a shorthand way of expressing it, without stating it explicitly.

Note: I've edited your comment to show you how to use the brackets \ ( \ ) around the math code to make the formulas display correctly. Staff · 2 years, 10 months ago

He just used the substitution you proposed, just in a disguised way. And you're right. Wolfram|Alpha agrees.

Calvin made a boo-boo · 2 years, 10 months ago

\sin{x} = $$\sin{x}$$

\sin^2{x}= $$\sin^2{x}$$

\theta = $$\theta$$

Use $$\(...$$) for latex. (... replace with the code, ex. sin{x}) · 2 years, 10 months ago

Thx...that was my first comment,so couldnt use it properly...p.s how he took d \sin{x} in place of dx · 2 years, 10 months ago

Eh just think of it as a one big blob (a chunk of "stuff"). He's integrating $$\sin{(x)}$$ with respect to $$\sin{(x)}$$. To make sense of this, he really used the substitution: Let $$t=\sin{(x)}$$. Then $$dt=\cos{(x)}dx$$. Then your integral will look like

$$\int{tdt}$$

Now think of $$t$$ as a blob of what it originally was ($$\sin{(x)}$$). If you do so, you end up with

$$\int{\sin{(x)}d(\sin{(x)}})$$

-see? Pretty clever, eh?

And yes you do end up with $$\frac{t^2}{2}+C$$, or, $$\frac{1}{2}\sin^2{x}+C$$.

P.S.

If you wonder why he did it, sometimes in problems its easy to see. If you notice, the stuff after $$d$$ is just the stuff he took out - integrated. This is so that to undo the $$d"stuff"$$, when you differentiate the "stuff" it will get back into the argument and $$d$$ will take its original form. This is a bit a wacko of an explanation but you can probably see it better from the $$t$$ substitution.

Hope this helped.

Cheers, · 2 years, 10 months ago

I understood ur explanation and logics but ur english is quite tough for me :D ... thx again · 2 years, 10 months ago

Glad to help :D · 2 years, 10 months ago

Thx for the replies...and btw I cant use codes... e.g \sin{x} ... doesnt work :p

Edit: Use $$\sin {x}$$. You can edit this comment (click the pencil in the top right), to see the actual code used. · 2 years, 10 months ago

For further guide click on "Formatting guide" below your reply interface. Two quick shortcuts they don't include: \frac{TOP}{BOTTOM}=$$\frac{TOP}{BOTTOM}$$; \int {xdx} =$$\int {xdx}$$.

Cool?

\pi =$$\pi$$

\infty = $$\infty$$

\Rightarrow = $$\Rightarrow$$

When raising to a power that includes more than two characters, encase the exponent with {...}. Ex.

\e^-x=$$e^-1x$$. \e^{-x}=$$e^{-x}$$

P.S. - double enter to skip a line · 2 years, 10 months ago

Like I said, encase all Latex code in \ ( ... \ ) (without the spaces). The "..." is the stuff you put in, like the \sin{x}. · 2 years, 10 months ago

(Y) ... · 2 years, 10 months ago

@hasan kassim @Krishna Jha Can you add this to the Wiki page of a suitable skill in Integration by Parts? Staff · 2 years, 9 months ago

How do I do that sir? · 2 years, 9 months ago

@Krishna Jha Can you add this to Integration by Parts, so that others can be kept aware of this common mistake? Thanks! Staff · 2 years, 9 months ago

OK.. But how to do that sir? Is this fine?? · 2 years, 9 months ago

bro,, this is indefinite integral,, several functions infact infinite functions all differing by a constant give us the same derivative,, · 2 years, 10 months ago

The difference between the primitives of a function is a constant.

Let $$F ' (x) = G ' (x) = f(x)$$.

Then $$F(x) = G(x) + C$$. · 2 years, 10 months ago

You cannot take a number that needs to be integrated out of the integration · 2 years, 10 months ago

U can't use integration by parts for integrating two same functions · 2 years, 10 months ago