Integrate $\frac{1}{\sqrt{(3-x)(x-4)}}$ to Get $-\pi$

  • Context: MHB 
  • Thread starter Thread starter Sherlock1
  • Start date Start date
  • Tags Tags
    Integration
Click For Summary

Discussion Overview

The discussion revolves around the integral $\displaystyle I = \int_{3}^{4}\frac{1}{\sqrt{(3-x)(x-4)}}\;{dx}$ and the apparent discrepancy in its evaluation, with some participants asserting it equals $\pi$ while others suggest it equals $-\pi$. The conversation explores various methods of integration, including substitutions and the use of the Beta function, and examines the implications of negative results in the context of definite integrals.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the integral evaluates to $\pi$, while others argue it evaluates to $-\pi$.
  • One participant suggests using the substitution $t = 3-x$, leading to the expression $-\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}$, which they claim equals $-\pi$.
  • Another participant mentions the use of the Beta function to evaluate the integral, arriving at $-\pi$ as well.
  • Concerns are raised about the validity of the negative result, with one participant noting that integrating a positive function over a positive interval should not yield a negative value.
  • There is a discussion about the implications of complex numbers when evaluating the square roots involved in the integral, with suggestions to rewrite the integral to avoid complex roots.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the value of the integral, with multiple competing views remaining regarding whether it equals $\pi$ or $-\pi$. The discussion highlights confusion and uncertainty about the methods used and the results obtained.

Contextual Notes

Participants express uncertainty about the handling of square roots in the integral, particularly regarding the signs of the terms involved and the implications of using complex numbers. There are unresolved questions about the proper interpretation of the integral's limits and the transformations applied.

Sherlock1
Messages
38
Reaction score
0
Consider $\displaystyle I = \displaystyle \int_{3}^{4}\frac{1}{\sqrt{(3-x)(x-4)}}\;{dx}.$ It can be shown that this is $\pi$.

Let $t = 3-x$ for the integral and you get $\displaystyle I = -\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}.$

However, the latter integral is equal to $-\pi$. Where is the negative from?
 
Last edited:
Physics news on Phys.org
Re: Something really weird

Sherlock said:
Consider $\displaystyle I = \int_{3}^{4}(3-x)^{-\frac{1}{2}}(x-4)^{ - \frac{1}{2}}\;{dx}.$ It can be shown that this is $\pi$.

Let $t = 3-x$ for the integral and you get $\displaystyle I = -\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}.$

However, the latter integral is equal to $-\pi$. Where is the negative from?

Hi Sherlock, :)

Both integrals should be equal to \(-\pi\). Can you write down your solution for the first integral, maybe you are missing a minus sign somewhere.

>>Integral 1<<

>>Integral 2<<

Kind Regards,
Sudharaka.
 
Re: Something really weird

Sudharaka said:
Hi Sherlock, :)

Both integrals should be equal to \(-\pi\). Can you write down your solution for the first integral, maybe you are missing a minus sign somewhere.

>>Integral 1<<

>>Integral 2<<

Kind Regards,
Sudharaka.
Hello, Sudharaka. I did plug that in Wolfram but it gave me this! (Giggle)
 
Re: Something really weird

Sherlock said:
Hello, Sudharaka. I did plug that in Wolfram but it gave me this! (Giggle)

I don't understand what is happening with wolfram, but you can solve this integral using the Beta function.

\[I = \int_{3}^{4}(3-x)^{-\frac{1}{2}}(x-4)^{ - \frac{1}{2}}\;{dx}\]

Substitute, \(t=x-3\) and we get,

\begin{eqnarray}

I &=& \int_{3}^{4}(3-x)^{-\frac{1}{2}}(x-4)^{ - \frac{1}{2}}\;{dx}\\

&=&-\int_{0}^{1}t^{-\frac{1}{2}}(1-t)^{-\frac{1}{2}}\,dt\\

&=&-\beta\left(\frac{1}{2},\frac{1}{2}\right)\\

\end{eqnarray}

Since, \(\beta(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}\!\) we have,

\[I=-\frac{\Gamma\left(\frac{1}{2}\right)\,\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)}=-\pi\]
 
Re: Something really weird

That's how I did it as well. But I still don't understand the negative anomaly.

Apparently $\displaystyle I = \int_{3}^{4}\bigg[(3-x)(x-4)\bigg]^{-\frac{1}{2}}\;{dx} = -\int_{3}^{4}(3-x)^{-\frac{1}{2}}(x-4)^{ - \frac{1}{2}}\;{dx}?$
 
Re: Something really weird

Sherlock said:
Consider $\displaystyle I = \int_{3}^{4}(3-x)^{-\frac{1}{2}}(x-4)^{ - \frac{1}{2}}\;{dx}.$ It can be shown that this is $\pi$.

No. Refer my previous post and you'll and see that it's equal to \(-\pi\).

Sherlock said:
That's how I did it as well. But I still don't understand the negative anomaly.

Apparently $\displaystyle I = \int_{3}^{4}\bigg[(3-x)(x-4)\bigg]^{-\frac{1}{2}}\;{dx} = -\int_{3}^{4}(3-x)^{-\frac{1}{2}}(x-4)^{ - \frac{1}{2}}\;{dx}?$

Can you please show how you calculated the second integral?
 
Re: Something really weird

Sudharaka said:
No. Refer my previous post and you'll and see that it's equal to \(-\pi\).
Can you please show how you calculated the second integral?

Sorry, the first post should have read:

Sherlock said:
Consider $\displaystyle \int_{3}^{4}\bigg[(3-x)(x-4)\bigg]^{-\frac{1}{2}}\;{dx}.$ It can be shown that this is $\pi$.

Let $t = 3-x$ for the integral and you get $\displaystyle I = -\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}.$

However, the latter integral is equal to $-\pi$. Where is the negative from?
Wolfram agrees that the first integral is $\pi$ (see my link above).

But as you have seen the second integral is $-\pi$. Why is this?
 
Re: Something really weird

Sherlock said:
Sorry, the first post should have read:

Wolfram agrees that the first integral is $\pi$ (see my link above).

But as you have seen the second integral is $-\pi$. Why is this?

I don't know why wolfram gives different answers when the problem is stated in different ways. Maybe some other member will be able to enlighten you as to why wolfram behaves that way. However we have obtained that,

\[\int_{3}^{4}(3-x)^{-\frac{1}{2}}(x-4)^{ - \frac{1}{2}}\;{dx}=-\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}\,dt=-\pi\]
 
Re: Something really weird

Sudharaka said:
I don't know why wolfram gives different answers when the problem is stated in different ways. Maybe some other member will be able to enlighten you as to why wolfram behaves that way. However we have obtained that,

\[\int_{3}^{4}(3-x)^{-\frac{1}{2}}(x-4)^{ - \frac{1}{2}}\;{dx}=-\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}\,dt=-\pi\]
You still don't get me. Let me illustrate what I mean by way of a solution:

Let $x = \frac{7}{2}+t$ then we have:$\displaystyle \begin{aligned} I & = \int_{3}^{4}\frac{1}{\sqrt{(3-x)(x-4)}}\;{dx} = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{\sqrt{(\frac{1}{2})^2-t^2}}\;{dt} = 2 \int_{0}^{\frac{1}{2}}\frac{1}{\sqrt{(\frac{1}{2})^2-t^2}}\;{dt} \\& = 2\sin^{-1}(2t)\bigg|_{0}^{1/2} = 2\sin^{-1}(1)-2\sin^{-1}(0) = \pi. \end{aligned}$

But if we do it the beta function way like you have done, we get $-\pi$.

I hope you understand what I mean now. This is driving me insane!
 
  • #10
Re: Something really weird

Here is the full version of Mathematica's solution.

https://www.physicsforums.com/attachments/460
 
  • #11
Re: Something really weird

dwsmith said:
Here is the full version of Mathematica's solution.

https://www.physicsforums.com/attachments/460
What does it give when you enter $\displaystyle \int_{3}^{4}\frac{1}{\sqrt{(3-x)(x-4)}}\;{dx}$ (both terms under the radical), though? Thanks.
 
  • #12
Re: Something really weird

Sherlock said:
What does it give when you enter $\displaystyle \int_{3}^{4}\frac{1}{\sqrt{(3-x)(x-4)}}\;{dx}$ (both terms under the radical), though? Thanks.

For that integral it, gave me pi whereas the other two gave the set containing -pi.
 
  • #13
Sherlock said:
Consider $\displaystyle I = \displaystyle \int_{3}^{4}\frac{1}{\sqrt{(3-x)(x-4)}}\;{dx}.$ It can be shown that this is $\pi$.

Let $t = 3-x$ for the integral and you get $\displaystyle I = -\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}.$

However, the latter integral is equal to $-\pi$. Where is the negative from?
The notation $\sqrt x$ (for a positive number $x$) by definition means the positive square root. If you integrate a positive function over a (positively oriented) interval, you can never get a negative answer. So if the answer comes out as $-\pi$, you know that something has gone wrong somewhere.

In this case, something is odd about the integral $-\displaystyle\int_{0}^{-1}t^{-\frac{1}{2}}(-1-t)^{-\frac{1}{2}}\,dt$, because $t$ and $-1-t$ are both negative when $-1<t<0$, and therefore do not have real square roots. But their product is positive, so you can write the integral as $-\displaystyle\int_{0}^{-1}\bigl(t(-1-t)\bigr)^{-\frac{1}{2}}\,dt = \int_{-1}^{0}\bigl(t(-1-t)\bigr)^{-\frac{1}{2}}\,dt$. This should come out as $+\pi$.
 
  • #14
Thanks. Could you explain bit more as to how inverting the limits voids the negative, please? I can't see it.

I know it cancels the immediate one before the integral, but in the end we still end-up with $-\pi$? Thanks!
 
  • #15
Sherlock said:
Thanks. Could you explain bit more as to how inverting the limits voids the negative, please? I can't see it.

I know it cancels the immediate one before the integral, but in the end we still end-up with $-\pi$? Thanks!
As far as I can see, the mistake comes from mishandling the expression $1/\sqrt{(3-x)(x-4)}$. If you write it as $(3-x)^{-1/2}(x-4)^{-1/2}$, then (in the interval $3<x<4$ over which you are integrating) that is the product of two complex numbers. Both $3-x$ and $x-4$ are negative in that interval, so each of them has two purely-imaginary square roots and there is no reliable way to select one of those roots rather than the other.

What you should do, if you want to transform the integral into a beta-function form, is to re-write $1/\sqrt{(3-x)(x-4)}$ as $1/\sqrt{(x-3)(4-x)} = (x-3)^{-1/2}(4-x)^{-1/2}$. That is now the product of two positive numbers and you can make the substitution $t=x-3$ to get the integral in the form $\displaystyle\int_0^1t^{-1/2}(1-t)^{-1/2}\,dt = \beta(1/2,1/2).$
 
  • #16
I have fallen into >>this trap<< a couple of times yet never learns form the mistake. (Headbang)

@Opalg: Thank you very much for your valuable insight into this problem. :)
 

Similar threads

High School I need to check if I am right solving this integral
  • · Replies 2 ·
Replies
2
Views
2K
High School Having trouble deriving the volume of an elliptical ring toroid
  • · Replies 4 ·
Replies
4
Views
4K
Undergrad Substitution in a definite integral
  • · Replies 6 ·
Replies
6
Views
3K
MHB How to Evaluate the Integral of an Even Trig Function in a Radical?
  • · Replies 9 ·
Replies
9
Views
1K
High School Trigonometric substitution, a case I'd like to share
  • · Replies 3 ·
Replies
3
Views
3K
Graduate A discrete version of the normal distribution
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Having trouble understanding a seemingly simple u-substitution
  • · Replies 5 ·
Replies
5
Views
2K
High School Question about the limits of integration in a change of variables
  • · Replies 2 ·
Replies
2
Views
3K
High School Integration by Parts, an introduction I get confused with
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad Express the limit as a definite integral
  • · Replies 16 ·
Replies
16
Views
4K