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According to many common definitions of an integral (for example, as the limit of Riemannian sums), the expression

∫ x d x ” id=”MathJax-Element-6-Frame” role=”presentation” tabindex=”0″>

$\int \frac{x}{dx}$is meaningless.

Roughly speaking, an integral is a sum over an infinity of zeros, expressed as the limit of the sum of values of the function times small intervals, as the intervals go to zero in size. In the integral

∫ a b f ( x ) d x ” id=”MathJax-Element-7-Frame” role=”presentation” tabindex=”0″>

${\int}_{a}^{b}f(x)dx$, the

d x ” id=”MathJax-Element-8-Frame” role=”presentation” tabindex=”0″>

$dx$represents the small, zero-trending intervals.

One version of this as a sum would be

∫ a b f ( x ) d x = lim N → ∞ ∑ i = 0 N f ( a i ) Δ x i ” id=”MathJax-Element-9-Frame” role=”presentation” tabindex=”0″>

${\int}_{a}^{b}f(x)dx=\underset{N\to \mathrm{\infty}}{lim}\sum _{i=0}^{N}f({a}_{i})\mathrm{\Delta}{x}_{i}$where

a i = a + i N ( b − a ) , Δ x i = 1 N ” id=”MathJax-Element-10-Frame” role=”presentation” tabindex=”0″>

${a}_{i}=a+\frac{i}{N}(b-a),\mathrm{\Delta}{x}_{i}=\frac{1}{N}$. There are other versions that work as well. The key being that as the limit goes to infinity, the size of the largest interval,

m a x ( Δ x i ) ” id=”MathJax-Element-11-Frame” role=”presentation” tabindex=”0″>

$max(\mathrm{\Delta}{x}_{i})$goes to zero.

The strict interpretation, thereof, of

∫ x d x ” id=”MathJax-Element-12-Frame” role=”presentation” tabindex=”0″>

$\int \frac{x}{dx}$would be the infinite sum of terms which go to infinity, not terms that go to zero. As such, it would be divergent, and wouldn’t exist.

Let us investigate this from a nonstandardpoint. Let I=[a,b] be a closed and bounded interval of real numbers, where

a ” id=”MathJax-Element-24-Frame” role=”presentation” tabindex=”0″>

$a$and

b ” id=”MathJax-Element-25-Frame” role=”presentation” tabindex=”0″>

$b$are real numbers with

b > a ” id=”MathJax-Element-26-Frame” role=”presentation” tabindex=”0″>

$b>a$, and let

ν ” id=”MathJax-Element-27-Frame” role=”presentation” tabindex=”0″>

$\nu $be an infinite hyperinteger. Let

d x = b − a ν ” id=”MathJax-Element-28-Frame” role=”presentation” tabindex=”0″>

$dx=\frac{b-a}{\nu}$. Note that

d x ” id=”MathJax-Element-29-Frame” role=”presentation” tabindex=”0″>

$dx$is a positive infinitesimal hyperreal number. For each non-negative hyperinteger

j ≤ ν ” id=”MathJax-Element-30-Frame” role=”presentation” tabindex=”0″>

$j\le \nu $, let

x j = a + j d x ” id=”MathJax-Element-31-Frame” role=”presentation” tabindex=”0″>

${x}_{j}=a+jdx$, so that the nonstandard expression for the “Left Riemann Sum” corresponding to the kind of integral you ask us to compute is the following:

∑ j = 0 ν x j d x = ∑ j = 0 ν ( a + j d x ) d x = ∑ j = 0 ν ( a d x + j ) = ( ν + 1 ) a d x + ( ν + 1 ) ν 2 = ( ν + 1 ) ( a d x + ν 2 ) = ( ν + 1 ) ( 2 a + ν d x 2 d x ) = 1 2 ( ν + 1 ) ( 2 a + ν d x b − a ν ) = 1 2 ( ν + 1 ) ( 2 a + ν d x b − a ν ) = ν ( ν + 1 ) 2 ( 2 a + b − a b − a ) = ν ( ν + 1 ) 2 ( 2 a b − a + 1 ) , ” id=”MathJax-Element-32-Frame” role=”presentation” tabindex=”0″>

$\begin{array}{rcl}\sum _{j=0}^{\nu}\frac{{x}_{j}}{dx}& =& \sum _{j=0}^{\nu}\frac{(a+jdx)}{dx}\\ & =& \sum _{j=0}^{\nu}(\frac{a}{dx}+j)\\ & =& (\nu +1)\frac{a}{dx}+\frac{(\nu +1)\nu}{2}\\ & =& (\nu +1)(\frac{a}{dx}+\frac{\nu}{2})\\ & =& (\nu +1)\left(\frac{2a+\nu dx}{2dx}\right)\\ & =& \frac{1}{2}(\nu +1)\left(\frac{2a+\nu dx}{\frac{b-a}{\nu}}\right)\\ & =& \frac{1}{2}(\nu +1)\left(\frac{2a+\nu dx}{\frac{b-a}{\nu}}\right)\\ & =& \frac{\nu (\nu +1)}{2}\left(\frac{2a+b-a}{b-a}\right)\\ & =& \frac{\nu (\nu +1)}{2}(\frac{2a}{b-a}+1)\end{array},$It follows that

∑ j = 0 ν x j d x ” id=”MathJax-Element-33-Frame” role=”presentation” tabindex=”0″>

$\sum _{j=0}^{\nu}\frac{{x}_{j}}{dx}$is an infinite hyperreal number, for every infinite hyperinteger

ν ” id=”MathJax-Element-34-Frame” role=”presentation” tabindex=”0″>

$\nu $and every corresponding partition of the interval

I = [ a , b ] ” id=”MathJax-Element-35-Frame” role=”presentation” tabindex=”0″>

$I=[a,b]$. Consequently, if the integral expression

∫ a b x d x ” id=”MathJax-Element-36-Frame” role=”presentation” tabindex=”0″>

${\int}_{a}^{b}\frac{x}{dx}$is to be interpreted as having a “value”, then that value must be “lazy eight”, i.e.

∞ ” id=”MathJax-Element-37-Frame” role=”presentation” tabindex=”0″>

$\mathrm{\infty}$. We used only “Left Riemann Sums” for this calculation, so to complete the analysis, we would need to consider arbitrary partitions of the interval

I = [ a , b ] ” id=”MathJax-Element-38-Frame” role=”presentation” tabindex=”0″>

$I=[a,b]$, in which the interval widths need not be the same, i.e., not “regular” partitions. We could do a reasonable calculation for Lebesgue integral type interpretation of this integral expression, and similarly, we would obtain

∞ ” id=”MathJax-Element-39-Frame” role=”presentation” tabindex=”0″>

$\mathrm{\infty}$as the result. (I will leave that as an exercise for those with the patience to suss out the details…)

Now, this can be modified so as to compute this integral like expression over other measurable subsets, E, of the real line, and the result will be that if E has positive measure, then the corresponding integral will be infinite, and I expect that otherwise, the corresponding integral will be zero, although I have not tried to compute it yet.

Powerful continuous integration out of the box.

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It is the same as the answer to

x = ( 3 y + 1 ” id=”MathJax-Element-50-Frame” role=”presentation” tabindex=”0″>

$x=(3y+\frac{1}{}$That is: it is not a well-formed expression. It is meaningless.

We use the notation

∫ f ( x ) d x ” id=”MathJax-Element-51-Frame” role=”presentation” tabindex=”0″>

$\int f(x)dx$with any function

f ( x ) ” id=”MathJax-Element-52-Frame” role=”presentation” tabindex=”0″>

$f(x)$for an integral. The term

d x ” id=”MathJax-Element-53-Frame” role=”presentation” tabindex=”0″>

$dx$is part of the notation.

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If you want to be really technical, what you are calling an integral is actually an anti-derivative (also known as an indefinite integral). The term integral alone is often reserved for the definite integral (i.e. one that has a set over which the integration is carried out).

The anti-derivative of

x x ” id=”MathJax-Element-56-Frame” role=”presentation” tabindex=”0″>

${x}^{x}$certainly exists (at least over appropriately defined domains); however, the function that is the anti-derivative does not have a representation using Elementary functions.

You might as well create a new function called the TianZong function. (I really hope that I’ve correctly identified the name of the author of the question!) The definition of the TianZong function would be that, for any

x > 0 ” id=”MathJax-Element-57-Frame” role=”presentation” tabindex=”0″>

$x>0$, TianZong(x) has the property that the slope of the tangent line of this function at this point is given by

x x ” id=”MathJax-Element-58-Frame” role=”presentation” tabindex=”0″>

${x}^{x}$, and I’ll throw in that TianZong

( 1 ) = 0 ” id=”MathJax-Element-59-Frame” role=”presentation” tabindex=”0″>

$(1)=0$so that it is uniquely defined.

Now that we have the function uniquely defined, we can set out to determine how the function behaves. For example, you might care about the value of TianZong(2). We see that:TianZong

( x ) = ∫ 1 x y y d y ” id=”MathJax-Element-60-Frame” role=”presentation” tabindex=”0″>

$(x)={\int}_{1}^{x}{y}^{y}dy$for any

x > 0 ” id=”MathJax-Element-61-Frame” role=”presentation” tabindex=”0″>

$x>0$.

So we can write:

TianZong(2)= ∫ 1 2 x x d x ” id=”MathJax-Element-62-Frame” role=”presentation” tabindex=”0″>

$={\int}_{1}^{2}{x}^{x}dx$It’s easy to numerically compute this integral to arbitrary precision. I used a Monte Carlo estimate to find that TianZong(2)

≈ 2.05045 ” id=”MathJax-Element-63-Frame” role=”presentation” tabindex=”0″>

$\approx 2.05045$. I used Simpson’s rule with 1 million grid points to get:

TianZong(2)≈ 2.050446234535 ” id=”MathJax-Element-64-Frame” role=”presentation” tabindex=”0″>

$\approx 2.050446234535$I got the same result (to this precision) using 10 million grid points, so that’s probably the right answer (to this precision).

Now, you might object to this approach. You might take offense to the idea of solving a problem by essentially making up a new function that is defined to be the answer to the problem. But realize that this approach is taken all the time. The Lambert W function was created to solve equations with exponential terms and linear terms. The Bessel function was created to solve a certain kind of differential equation. The Gamma function (like the newly minted TianZong function) was created to be the answer to a certain integral.

When mathematicians get a problem, they try to solve it using tools that already exist. If they can’t do so, they try to prove that it can’t be solved using tools that already exist. If they can prove such a result (and if the problem is interesting enough that things like it will come up in other problems), they construct brand new tools that can solve the problem at hand and are ready to be applied to future related problems.

Greetings!

While I do believe Angad Kapoor has done justification to the mathematical manipulation involved, that is all I believe this is. A party trick, perhaps.

Mathematics may have an abstract concept often, but this fails to meet that criteria. An integral is the sum of infinitesimal area elements. And area of a rectangle(since it is Riemann integral) is length breadth. The function value is the height and dx is the breadth of the area element.

When the integral lacks a dx multiplied to a function, it ceases to have any meaning mathematically. Mathematically, this is sacrilege and blasphemy of the highest order. But yes, it is a cool trick, it is.

May the Force be with you.

Should I hire remote software developers from Turing.com?

It is so hard to hire strong engineers for my company in San Francisco.

As Dan Loewenherz says, this amounts to asking “Why is

1 / x ” id=”MathJax-Element-74-Frame” role=”presentation” tabindex=”0″>

$1/x$the derivative of

ln ( x ) ” id=”MathJax-Element-75-Frame” role=”presentation” tabindex=”0″>

$\mathrm{ln}(x)$?”. However, I think writing out limits is an unnecessarily complicating way to think about this.

The simplest way is by setting

y = ln ( x ) ” id=”MathJax-Element-76-Frame” role=”presentation” tabindex=”0″>

$y=\mathrm{ln}(x)$(equivalently,

x = e y ” id=”MathJax-Element-77-Frame” role=”presentation” tabindex=”0″>

$x={e}^{y}$), and then observing that

d ln ( x ) d x = d y d e y = 1 / ( d e y d y ) = 1 / ( e y ) = 1 / x ” id=”MathJax-Element-78-Frame” role=”presentation” tabindex=”0″>

$\frac{d\mathrm{ln}(x)}{dx}=\frac{dy}{d{e}^{y}}=1/(\frac{d{e}^{y}}{dy})=1/({e}^{y})=1/x$.

Thus, this is just a head-tilted way of looking at the fact that

e t ” id=”MathJax-Element-79-Frame” role=”presentation” tabindex=”0″>

${e}^{t}$is its own derivative.

—-

The question of course can be asked “Ok, but why ise t ” id=”MathJax-Element-80-Frame” role=”presentation” tabindex=”0″>

${e}^{t}$its own derivative?”. Essentially, this is (or should be taken as) the defining property of

e ” id=”MathJax-Element-81-Frame” role=”presentation” tabindex=”0″>

$e$; specifically,

e ” id=”MathJax-Element-82-Frame” role=”presentation” tabindex=”0″>

$e$is defined to be the base of the natural logarithm (that is, via

ln ( e ) = 1 ” id=”MathJax-Element-83-Frame” role=”presentation” tabindex=”0″>

$\mathrm{ln}(e)=1$), with the natural logarithm

ln ( x ) ” id=”MathJax-Element-84-Frame” role=”presentation” tabindex=”0″>

$\mathrm{ln}(x)$, in turn, being defined as the invariant ratio between

d x t d t ” id=”MathJax-Element-85-Frame” role=”presentation” tabindex=”0″>

$\frac{d{x}^{t}}{dt}$and

x t ” id=”MathJax-Element-86-Frame” role=”presentation” tabindex=”0″>

${x}^{t}$itself (equivalently, as the value of

d x t d t ” id=”MathJax-Element-87-Frame” role=”presentation” tabindex=”0″>

$\frac{d{x}^{t}}{dt}$at

t = 0 ” id=”MathJax-Element-88-Frame” role=”presentation” tabindex=”0″>

$t=0$). Thus, that exponentiation with base

e ” id=”MathJax-Element-89-Frame” role=”presentation” tabindex=”0″>

$e$is its own derivative is tautologous;

e ” id=”MathJax-Element-90-Frame” role=”presentation” tabindex=”0″>

$e$is defined to be the base with this property, that being its entire raison d’etre.

This understanding of the definition of the natural logarithm actually leads into another way of seeing the same argument presented above, if one wants to unify it with the general power rule for integration: Consider the integral

∫ 1 x x n − 1 d x ” id=”MathJax-Element-91-Frame” role=”presentation” tabindex=”0″>

${\int}_{1}^{x}{x}^{n-1}dx$. Let us refer to this as

h ( n , x ) ” id=”MathJax-Element-92-Frame” role=”presentation” tabindex=”0″>

$h(n,x)$; as is well known,

n h ( n , x ) = x n − 1 ” id=”MathJax-Element-93-Frame” role=”presentation” tabindex=”0″>

$nh(n,x)={x}^{n}-1$, which is the general “power rule” for integration. In particular, applying

d d n ” id=”MathJax-Element-94-Frame” role=”presentation” tabindex=”0″>

$\frac{d}{dn}$to both sides and then setting

n ” id=”MathJax-Element-95-Frame” role=”presentation” tabindex=”0″>

$n$to zero, the left side becomes

h ( 0 , x ) = ∫ 1 x x − 1 d x ” id=”MathJax-Element-96-Frame” role=”presentation” tabindex=”0″>

$h(0,x)={\int}_{1}^{x}{x}^{-1}dx$, while the right side, practically by definition (as given above), becomes

ln ( x ) ” id=”MathJax-Element-97-Frame” role=”presentation” tabindex=”0″>

$\mathrm{ln}(x)$, thus establishing that the integral of

x − 1 ” id=”MathJax-Element-98-Frame” role=”presentation” tabindex=”0″>

${x}^{-1}$is indeed

ln ( x ) ” id=”MathJax-Element-99-Frame” role=”presentation” tabindex=”0″>

$\mathrm{ln}(x)$The operation of definite integration acts on a function and returns a number; indefinite integration acts on a function and returns another function.

“dx” is an infinitesimal, or more sophisticatedly is a measure, but it is not a function. Integration on anything not a function isn’t a defined operation.

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Since integration of (1/dx) is int((1/dx)*dx) = int(1)

This can be thought of integration independent of dependent variable.

So we are just adding 1 infinite times, hence integration is infinite.

If we consider 1/dx a function, it would look like a straight line y=a at infinite distance ‘a’ from x axis, such that a*dx=1.

Based on our choice of dx and integration limit, we can get different values of our function, hence, integration. Eg. dx=0.1 and integration is from 0 to 10. It would mean we’re adding 1, 100 times. Since dx is much smaller than 0.1, our answer will be tending to infinite.

This kind of integration may not be a standard kind, but if we allow ourselves that there can exist a function that has a variable that is representing depth of our integration (i.e. close to actual definition…nes), we can make sense out of the integral.

This can be function that could check accuracy of computer program for integration.

Who knows, wether you had actual doubt or you were just trolling. But maths is not rigid guidelines ment to be followed by meatballs. It is our tautology towards reality. And even if we are not talking about reality, maths gives us power to convince our own realty and examine it’s facts and consequences.

You had a little doubt that seemingly had no meaning at first glance. Keep asking these doubts, but next time try to answer them by yourself. That will help a lot on deep level.

∫ d x = ∫ 1 d x = x + C ” id=”MathJax-Element-102-Frame” role=”presentation” tabindex=”0″>

$\int \mathrm{d}x=\int 1\mathrm{d}x=x+C$Actually, it’s not so proper to rewrite

1 ” id=”MathJax-Element-103-Frame” role=”presentation” tabindex=”0″>

$1$as

x 0 ” id=”MathJax-Element-104-Frame” role=”presentation” tabindex=”0″>

${x}^{0}$because domains are different for

f ( x ) = 1 ” id=”MathJax-Element-105-Frame” role=”presentation” tabindex=”0″>

$f(x)=1$and

f ( x ) = x 0 ” id=”MathJax-Element-106-Frame” role=”presentation” tabindex=”0″>

$f(x)={x}^{0}$.

- We can find
F ( x ) ” id=”MathJax-Element-107-Frame” role=”presentation” tabindex=”0″>

$F(x)$, which is one antiderivative of

f ( x ) = 1 ” id=”MathJax-Element-108-Frame” role=”presentation” tabindex=”0″>

$f(x)=1$. For example, we can suppose that

F ( x ) = x + 5 ” id=”MathJax-Element-109-Frame” role=”presentation” tabindex=”0″>

$F(x)=x+5$(because

F ′ ( x ) = ( x + 5 ) ′ = f ( x ) = 1 ” id=”MathJax-Element-110-Frame” role=”presentation” tabindex=”0″>

${F}^{\prime}(x)=(x+5{)}^{\prime}=f(x)=1$).

- Then plus a constant
C 1 ” id=”MathJax-Element-111-Frame” role=”presentation” tabindex=”0″>

${C}_{1}$, we’ve got

∫ d x = F ( x ) + C 1 = x + 5 + C 1 ” id=”MathJax-Element-112-Frame” role=”presentation” tabindex=”0″>

$\int \mathrm{d}x=F(x)+{C}_{1}=x+5+{C}_{1}$.

- Let
C = C 1 + 5 ” id=”MathJax-Element-113-Frame” role=”presentation” tabindex=”0″>

$C={C}_{1}+5$, then the answer is

∫ d x = x + C ” id=”MathJax-Element-114-Frame” role=”presentation” tabindex=”0″>

$\int \mathrm{d}x=x+C$.

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I wonder why quora have not merged this question with other same question with other lines as How one can integrate x^x? .

Now i will not give my own answer as many one have answered the question in very nice way like :-

By –Siddhant Grover

Hope it helps ! 🙂

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Let us start with basic division principle, half of 1 full thing i.e 1/2 = 0.5. In the same way if we divide 1 thing into 2 equal parts then each part will be equal to 0.5. So if we divide a thing, say a number divided into as many parts as possible then each part will be very tiny resulting in answers like 0.0000000……………………1. If we are able to divide a number into (hypothetically) infinite times then each part will be equal to ‘0’. This implies that any number divided by infinity gives Zero.

A n y n u m b e r I n f i n i t y ” id=”MathJax-Element-119-Frame” role=”presentation” tabindex=”0″>

$\frac{Anynumber}{Infinity}$= Zero.

Implying

A n y n u m b e r Z e r o ” id=”MathJax-Element-120-Frame” role=”presentation” tabindex=”0″>

$\frac{Anynumber}{Zero}$= Infinity.

Infinity is still undefined so

A n y n u m b e r Z e r o ” id=”MathJax-Element-121-Frame” role=”presentation” tabindex=”0″>

$\frac{Anynumber}{Zero}$= Undefined.

The arbitrary constant (Any number) can also be Zero.

Therefore

0 0 ” id=”MathJax-Element-122-Frame” role=”presentation” tabindex=”0″>

$\frac{0}{0}$= Undefined

This is why Zero and Zero doesn’t get cancelled to get 1.

Now coming to basic to basic principle of anti derivative i.e Integration. Derivative is nothing but slope of a function. Generally we deal with straight lines whose slope is constant. But in higher order equations such as quadratic and cubic equations, slope also changes with change in abscissa. But for a straight line parallel to X- axis slope will be zero. So derivative of Y = K in which Y is not a function of X , often called as constant function is Zero. Implying that Anti derivative i.e

∫ 0 d x ” id=”MathJax-Element-123-Frame” role=”presentation” tabindex=”0″>

$\int 0dx$is a constant which can be defined if proper limits are given but not undefined.

You can also check out following answers of mine to get clear understanding about limits and Integrals.

Pavan Chandaluri’s answer to Constant/infinity=0. Now, what is the reason we can’t say 0*infinity=constant? Is it because the limit is approaching the zero and infinity?Pavan Chandaluri’s answer to What are some tricks for understanding calculus that most professors don’t know, or don’t share with students?

You have to define what x to the power dx is, in order for me to solve, yet I wonder if you can come up with something remarkable, the integral you use will be a par of the definition needed.

obviously an integral of dx is x

I pause and consider as dx gets so small as to be 0, then x power 0 is 1, and so that option is simply out of bounds being an infinite sum of positives approaching 1.

So 1 over dx need be your integration path, but I cannot say, as you have not defined, so I could be completely wrong in understanding you.

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- We can find