Deriving the Integral via Leibniz Rule

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SUMMARY

The discussion centers on the evaluation of the integral \(\int_0^1\frac{x-1}{\ln{x}} dx\) using the Leibniz rule. The function \(\Phi(\alpha)=\int_0^1\frac{x^{\alpha}-1}{\ln{x}} dx\) is introduced, with the derivative \(\Phi'(\alpha)=\int_0^1\frac{x^{\alpha}\ln{x}}{\ln{x}} dx\) leading to the conclusion that \(\Phi'(\alpha)=\frac{1}{\alpha+1}\). A misunderstanding regarding the differentiation variable was clarified, confirming the correctness of the derivative calculation.

PREREQUISITES
  • Understanding of integral calculus, specifically the Leibniz rule.
  • Familiarity with logarithmic functions and their properties.
  • Knowledge of differentiation techniques, particularly with respect to parameters.
  • Basic comprehension of exponential functions and their logarithmic forms.
NEXT STEPS
  • Study the application of the Leibniz rule in evaluating parameter-dependent integrals.
  • Learn about the properties of logarithmic and exponential functions in calculus.
  • Explore advanced differentiation techniques, including differentiation under the integral sign.
  • Investigate the implications of the Fundamental Theorem of Calculus in relation to parameterized integrals.
USEFUL FOR

Students and educators in mathematics, particularly those focusing on calculus and integral evaluation techniques, as well as anyone interested in the application of the Leibniz rule in mathematical analysis.

dirk_mec1
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Homework Statement


[tex] \int_0^1\frac{x-1}{\ln{x}} dx[/tex]

Homework Equations


[tex] \Phi(\alpha)=\int_0^1\frac{x^{\alpha}-1}{\ln{x}} dx[/tex]

The Attempt at a Solution


In the answers they say:[tex] \Phi '(\alpha)=\int_0^1\frac{x^{\alpha}\ln{x}}{\ln{x}} dx=\frac{1}{\alpha+1}[/tex]but the derative is wrong, right? I don't understand how they calculated the derative...
 
Physics news on Phys.org
The derivative is d/d(alpha). x^alpha=e^(log(x)*alpha). Actually, it is right.
 
Dick said:
The derivative is d/d(alpha). x^alpha=e^(log(x)*alpha). Actually, it is right.

You're right I accendentally differentiated w.r.t x, thanks Dick.
 

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