# Thread: Integrate with respect to u

1. ## Integrate with respect to u

Hi, I am having difficulty integrating the following with respect to u:

$\displaystyle e^{\frac{1}{2}u^2 - \frac{3}{2}\ln u + c}$

Where c is a constant

2. Separate the integrals, and then use integration by parts.

3. EDIT: Wrong post... I thought that the solution could easily be obtained... my bad

4. Originally Posted by Unknown008
$\displaystyle \int e^{f(x)} dx = \frac{e^{f(x)}}{f'(x)} +c$
I think this is not correct. If we try to differentiate the right hand side, we don't get the function $\displaystyle e^{f(x)}$. For example, there is no closed form for $\displaystyle \int e^{x^2}dx$.
To answer the question, note that $\displaystyle e^{\frac{1}{2}u^2 - \frac{3}{2}\ln u + c} = e^c.e^{\frac{1}{2}u^2}.u^{-3/2}$. Now, change your variable and continue (you may need to integrate by parts)
But it gets very messy!
http://www.wolframalpha.com/input/?i=integrate+e^(0.5*x^2-1.5*ln(x)%2Bc)dx++
The answer is not encouraging :S

I think this is not correct. If we try to differentiate the right hand side, we don't get the function $\displaystyle e^{f(x)}$. For example, there is no closed form for $\displaystyle \int e^{x^2}dx$.
To answer the question, note that $\displaystyle e^{\frac{1}{2}u^2 - \frac{3}{2}\ln u + c} = e^c.e^{\frac{1}{2}u^2}.u^{-3/2}$. Now, change your variable and continue (you may need to integrate by parts)
But it gets very messy!
http://www.wolframalpha.com/input/?i=integrate+e^(0.5*x^2-1.5*ln(x)%2Bc)dx++
The answer is not encouraging :S
Is that so? Sorry, I thought that it worked like the exponential functions I've been working with

I tried on wolfram... it's the first time I see those symbols!

http://www.wolframalpha.com/input/?i=\int+e^{\frac{1}{2}u^2+-+\frac{3}{2}\ln+u+%2B+c}+du

6. Originally Posted by Anne05
Hi, I am having difficulty integrating the following with respect to u:

$\displaystyle e^{\frac{1}{2}u^2 - \frac{3}{2}\ln u + c}$

Where c is a constant
This is the same as $\displaystyle Ce^{\frac{u^2}{2}u^{\frac{3}{2}}$ with $\displaystyle C= e^c$. If we let $\displaystyle x= \frac{u^2}{2}$, then $\displaystyle dx= u du$ and $\displaystyle u^{\frac{3}{2}}du= (udu)(u^{\frac{1}{2}}= (2x)^{\frac{1}{4}}dx$ and $\displaystyle \int e^{\frac{u^2}{2}}= 2^{\frac{1}{4}}\int x^{\frac{1}{4}}e^x dx$
I don't believe that can be integrated in terms of elementary functions.

7. Originally Posted by Unknown008
I tried on wolfram... it's the first time I see those symbols!

http://www.wolframalpha.com/input/?i=\int+e^{\frac{1}{2}u^2+-+\frac{3}{2}\ln+u+%2B+c}+du
Well the only mysterious symbol is $\displaystyle \Gamma$ which is known as the Gamma function. It can be explained as a generalization of the factorial operator. It works on all real numbers. It is defined as integral.
check it: Gamma Function -- from Wolfram MathWorld

8. Yes, I was referring to that symbol, Gamma and another one, like 'O' in the series expansion of the integral. Thanks for the link! I don't think I'll understand these this year, maybe next year