Integration by Parts of <C> i just cannot do

AI Thread Summary
The discussion focuses on the step-by-step process of performing integration by parts for the integral involving the variable <C>. The integral I'_{1} is defined as I'_{1}(\alpha)=∫_{0}^{+\infty} C e^{-\alpha C^{2}} dC, which evaluates to 1/(2α). The integral I'_{3} is derived from I'_{1} by differentiating with respect to α, resulting in I'_{3}(\alpha)=∫_{0}^{+\infty} C^{3} e^{-\alpha C^{2}} dC equating to 1/(2α²). A correction was made regarding the inclusion of π in the final answer. The thread emphasizes the importance of understanding these integrals for exam preparation.
Rachael_Victoria
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Can anyone outline, and this is a rather large request, the step by step integration by parts for <C>? This is not a homework question but more something i need to be able to do on tuesday for my final, and have been trying to do for two days.
 

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Your answer is incorrect.

Here's how it's done.

Knowing this integral's value

I&#039;_{1}(\alpha)=:\int_{0}^{+\infty} C e^{-\alpha C^{2}} \ dC =\frac{1}{2\alpha} (1)

,one sees immediately that

I&#039;_{3}(\alpha)=:\int_{0}^{+\infty} C^{3} e^{-\alpha C^{2}} \ dC=-\frac{d}{d\alpha}I&#039;_{1}=\frac{1}{2\alpha^{2}} (2)

and now plug in (2) the expression for '\alpha'.

Daniel.
 
Thank you, and you're right, i forgot to type the pi in the denominator for the answer.
 
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