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 July 25th, 2012, 10:43 AM #1 Member   Joined: Apr 2010 Posts: 43 Thanks: 0 integrable functions Is it possible to prove that for every integrable function there is only one solution(for example$\int\frac{1}{x}\ dx=\ln x + C$?(I dont mean with different constants, I mean is it possible to prove, that there is not any other function, which would solve that problem).
 July 25th, 2012, 11:15 AM #2 Senior Member   Joined: Jul 2011 Posts: 118 Thanks: 0 Re: integrable functions $\int f(x)\,dx=F(x)\Leftrightarrow F#39;(x)=f(x)$ Let's assume that we have two solutions $F_1(x),F_2(x)$, then $$F_1(x)-F_2(x)$'=F_1'(x)-F_2#39;(x)=f(x)-f(x)=0\Rightarrow F_1(x)-F_2(x)=const$ So if $F(x)$ is the solution, another solution is only $F(x)+C$
 July 25th, 2012, 03:53 PM #3 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: integrable functions If i remember correctly some trigonometric functions have at least 2 different solutions for the integral. Can't remember which ones... was it $tan^{-1}$ or $sec^{-1}$? I do remember being surprised about it and showing my calc professor years ago and he was speechless Maybe someone else can elaborate more...
 July 25th, 2012, 03:55 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: integrable functions I know there are cases where the anti-derivative may be expressed in different forms, but it can be shown these different forms are the same function.
July 25th, 2012, 04:17 PM   #5
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Re: integrable functions

Quote:
 Originally Posted by MarkFL I know there are cases where the anti-derivative may be expressed in different forms, but it can be shown these different forms are the same function.
Maybe an example of what MarkFL is talking about is

$\int cos x dx= sin x + c$

and

$\int cos x dx= \frac {1}{2}ie^{-ix} - \frac {1}{2}ie^{ix} + c$

But what i saw years ago was different and now i don't remember anything except i was genuinely surprised and professor didn't explain...

 July 25th, 2012, 04:28 PM #6 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: integrable functions Here is an example of multiple forms for the same integral, all differing by a constant:: http://en.wikipedia.org/wiki/Integral_o ... t_function
July 25th, 2012, 04:38 PM   #7
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Re: integrable functions

Quote:
 Originally Posted by The_Fool Here is an example of multiple forms for the same integral, all differing by a constant:: http://en.wikipedia.org/wiki/Integral_o ... t_function
Yes, nice example, but are all 3 forms in the wiki article exactly the same function in disguise?

 July 25th, 2012, 05:51 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: integrable functions I believe the cases I am thinking of derive from the following identities: $\operatorname{arsinh}(x)=\ln$$x+\sqrt{x^{2}+1}$$ \operatorname{arcosh}(x)=\ln$$x+\sqrt{x^{2}-1}$$\,;x\ge1 \operatorname{artanh}(x)=\frac{1}{2}\ln$$\frac{1+x }{1-x}$$\,;|x|<1 \operatorname{arcoth}(x)=\frac{1}{2}\ln$$\frac{x+1 }{x-1}$$\,;|x|>1 \operatorname{arsech}(x)=\ln$$\frac{1+\sqrt{1-x^{2}}}{x}$$\,;0 \operatorname{arcsch}(x)=\ln$$\frac{1}{x}+\frac{\s qrt{1+x^2}}{|x|}$$\,;x\ne0$
July 25th, 2012, 07:07 PM   #9
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Re: integrable functions

Quote:
Originally Posted by agentredlum
Quote:
 Originally Posted by The_Fool Here is an example of multiple forms for the same integral, all differing by a constant:: http://en.wikipedia.org/wiki/Integral_o ... t_function
Yes, nice example, but are all 3 forms in the wiki article exactly the same function in disguise?
Correct. Another example is this integral:
$\int \sqrt{1-x^{\small 2}}dx$
To integrate this you'd need to do a trigonometric substitution. However, you could pick either sin or cos. If you work it out twice, one for each, you'll arrive at what appears to be different integrals. However, they are actually the same as they only differ by a constant.

 July 26th, 2012, 02:48 AM #10 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: integrable functions Since the differentiation of any function is unique; integration of any function is also unique. If $\int f(x) dx= F_1(x) \text{ and also } \int f(x) dx = F_2(x)$ Then it is possible to show that $F_1(x)$ is a disguised $F_2(x)$.

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