Substitution of x in Integral

Question

Hi I was wondering how to do the following:

I have this integral: $$\int _ { - L / 2 } ^ { + L / 2 } w' ( x )\, d x$$

I know would like to normalize x with respect to L: $$\hat { x } = \frac { x } { L }$$

How do I substitute correctly ?

I know that: $$w' ( x ) = \frac { 1 } { L } w' (\hat { x })$$

but what about the $dx$?

Thanks!

EDIT: From the chain rule: $$ \frac{dw}{dx} = \frac{d\hat{x}}{dx} \frac{dw}{d\hat{x}} = \frac{1}{L} \frac{dw}{d\hat{x}} $$

see here: Change of variables in differential equation?

The differential $\mathrm d x$ transforms linearly, that is $\mathrm d x = \mathrm d L \hat x = L \mathrm d \hat x$. But are you sure about $w(x) = \frac 1 L w(\hat x)$? — k.stm, Dec 14 '18 at 15:54
@james you shall not assume that unless you know that $w$ satisfy the following relation. After substitution, $w(x)$ will trasform into $w(L \hat{x})$ — hyperkahler, Dec 14 '18 at 15:57
@james Note it is w'(x) and not w^{'}(x) that has quite bad output. — egreg, Dec 14 '18 at 18:20

Shubham Johri · Accepted Answer · 2018-12-14T18:15:17.807

1

Note. You know that $\displaystyle\frac{d[w(x)]}{dx}=w'(x)\implies d[w(x)]=w'(x)dx$

$\displaystyle\int_{-L/2}^{L/2}w'(x)dx=\int_{-L/2}^{L/2}d[w(x)]=w(L/2)-w(-L/2)$

Method $1$

$\hat x=\frac xL\implies x=L\hat x$

Since $x$ ranges from $-L/2\to L/2, \hat x=\frac xL$ ranges from $-1/2\to1/2$.

Substitute $x$ by $L\hat x$ wherever you find $x$ in the integrand.

$\displaystyle\int_{-L/2}^{L/2}w'(x)dx=\int_{-1/2}^{1/2}w'(L\hat x)d(L\hat x)=L\int_{-1/2}^{1/2}w'(L\hat x)d\hat x\because d(L\hat x)=Ld(\hat x)$

Method $2$

$\displaystyle w'(x)=\frac d{dx}[w(x)]=\frac d{dx}[w(y)]$, where $y=L\hat x=x$

$\displaystyle w'(x)=\frac d{dx}[w(y)]=\frac d{dy}[w(y)]\times \frac{dy}{dx}=w'(y)\times\frac{dy}{dx}=w'(y)\times\frac{dx}{dx}=w'(y)=w'(L\hat x)$

The error in your approach lies when you claim that $\displaystyle\frac{d[w(x)]}{d\hat x}=w'(\hat x)$. Since $\displaystyle x=L\hat x,\frac{d[w(x)]}{d\hat x}=\frac{d[w(L\hat x)]}{d\hat x}$

By the chain rule, for finding this expression, you have to differentiate it with respect to the inner function and multiply with the derivative of the inner function with respect to $\hat x$.

$\displaystyle\frac{d[w(L\hat x)]}{d\hat x}=\frac{d[w(L\hat x)]}{d(L\hat x)}\times\frac{d(L\hat x)}{d\hat x}=Lw'(L\hat x)\because w'(L\hat x)=\frac d{d(L\hat x)}[w(L\hat x)]$

edited Dec 14 '18 at 18:15

answered Dec 14 '18 at 15:59

Shubham Johri

17,659

Can't you go further with $w(L\hat{x})$ ? – james Dec 14 '18 at 16:05
Not unless we have some more information about $w(x)$ – Shubham Johri Dec 14 '18 at 16:06
Please have a look at my new edit. Thanks ! – james Dec 14 '18 at 16:09
Oh, sorry ! I made an error in my question. It is now fixed. Would you mind to have a second look ? – james Dec 14 '18 at 16:12
Note that $\displaystyle\frac{dw}{d\hat x}\ne w'(\hat x)$ – Shubham Johri Dec 14 '18 at 16:12
@ Yes, this is my error. Sorry. I fixed it in the question. – james Dec 14 '18 at 16:13
$\displaystyle\int_{-L/2}^{L/2}w'(x)dx=w(L/2)-w(-L/2)$. Anyhow, note that $\displaystyle w=w(x)=w(L\hat x)\implies w'(x)=\frac{d[w(x)]}{dx}=\frac{d[w(L\hat x)]}{d(L\hat x)}\times\frac{d(L\hat x)}{dx}=w'(L\hat x)\times\frac{dx}{dx}$ – Shubham Johri Dec 14 '18 at 16:20
Thank you very much and sorry to bother you, but I still don't quite see how the result should look like ? Maybe would you mind to extend your answer just a bit ? – james Dec 14 '18 at 16:22
Alright, I'll add some more details – Shubham Johri Dec 14 '18 at 16:22
Thank you very much ! What happens if I have $w'(x)^2$ ? – james Dec 14 '18 at 17:48
When you have $[w'(x)]^2$, the answer will not be $w(L/2)-w(-L/2)$, but the substitution will be done similarly, yielding the integral $\displaystyle L\int_{-1/2}^{1/2}[w'(L\hat x)]^2d\hat x$ – Shubham Johri Dec 14 '18 at 17:50
Just remember to write $L\hat x$ wherever you see $x$, obtain the limits of the new variable and simplify – Shubham Johri Dec 14 '18 at 17:53
So, you say that the calculation carried out , provided by the link in my question is wrong ? – james Dec 14 '18 at 18:04
The link you provided says the same thing as I do – Shubham Johri Dec 14 '18 at 18:09
According to the accepted answer in the link, $\displaystyle f(L \hat{x}) =w''(L\hat x)=\frac1{L^2} \frac{d^2}{d\hat{x}^2} w(L\hat{x})=\frac{d^2}{d[L\hat{x}]^2} w(L\hat{x})$. I reached a similar conclusion in my answer (note the last line) – Shubham Johri Dec 14 '18 at 18:12
Ah, yes, you are right. Thanks a lot for your patience and help ! :) – james Dec 14 '18 at 18:18

egreg · Answer 2 · 2018-12-14T21:01:49.823

1

If you substitute $\hat{x}=x/L$, then $d\hat{x}=L^{-1}\,dx$ and the integral becomes $$ \int_{-1/2}^{1/2} Lw'(L\hat{x})\,d\hat{x} $$ Not really a progress, because you need to apply the chain rule to show that, if $u(\hat{x})=w(L\hat{x})$, then $$ u'(\hat{x})=Lw'(L\hat{x}) $$ and so the integral can be rewritten as $$ \int_{-1/2}^{1/2} u'(\hat{x})\,d\hat{x}=u(1/2)-u(-1/2)=w(L/2)-w(-L/2) $$ which was directly clear from the beginning with the fundamental theorem of calculus.

edited Dec 14 '18 at 21:01

answered Dec 14 '18 at 18:31

egreg

238,574

@ShubhamJohri Yes, you're right. Fixed. – egreg Dec 14 '18 at 21:02

Substitution of x in Integral

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