Logarithms & Integrals

Calculus Level 5

$\large I=\int^\frac \pi 2_0 e^{\ln\left ( \cos x \, \frac{\mathrm{d}\cos x}{\mathrm{d}x}\right ) } \mathrm{d}x$

If complex logarithms and complex exponentials are not allowed, what is $I$ equal to?

This question is derived from the Logarithm Definite Integral question.

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-\frac{1}{2}

None of these

\frac{1}{2}

-1

1

2 solutions

Utsav Banerjee
May 21, 2015

$\large e^{\ln\left(\cos x \times \frac{\mathrm{d}(\cos x)}{\mathrm{d}x}\right)}=e^{\ln(\cos x \times (-\sin x))}=e^{\ln(-\frac{1}{2} \sin (2x))}$

Since $\sin (2x)>0$ in the interval $x \in (0,\frac{\pi}{2})$ , the logarithm will not be defined in this interval, and the integral $I$ will not exist.

Note that $e^{\ln(x)}=x$ only in the domain of $\ln(x)$ , that is, only for $x>0$ .

How would you explain this??

When I wolframed it to crosscheck I got $\large{\frac{-1}{2}}$

Harshvardhan Mehta - 6 years ago

I think for normal logarithmic function, that would be undefined. Wolfram alpha probably used complex analysis for the logarithm. More to read : Link

Dawar Husain - 6 years ago

nopes it aint complex analysis.. even by solving it normally we get the answer $\displaystyle{\frac{-1}{2}}$

Harshvardhan Mehta - 6 years ago

How did you get the horizontal line in your comment?????

Raghav Vaidyanathan - 6 years ago

I also you cheater

Shashank Rustagi - 6 years ago

Thanks. Those who answered -1/2 have been marked correct. The problem has been edited and the correct answer is now "none of these".

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the “dot dot dot” menu in the lower right corner. This will notify the problem creator who can fix the issues.

Calvin Lin Staff - 6 years ago

Ahhh! you got me! I forgot to check out the limits.+1

Aditya Kumar - 6 years ago

But if x>0 then exp {ln(-x)}=exp {ln x + i(2n+1)(pi)} [where n is an integer.] =x exp[ i(2n+1)(pi)]= -x.

Anirban Karan - 6 years ago

However, if I consider complex numbers coming into action, this problems can be solved. What say?

A Former Brilliant Member - 6 years ago

Thanks. Those who answered -1/2 have been marked correct. The problem has been edited and the correct answer is now "none of these".

Calvin Lin Staff - 6 years ago

Without using complex exponentials we can obtain $-\dfrac{1}{2}$ as the answer if we simply rewrite $-\sin{x}$ as $\sin{(-x)}$ . Any thoughts?

Akeel Howell - 4 years, 2 months ago

How ? post your solution please.

Harshvardhan Mehta - 4 years ago

I can't write a solution as I got the question wrong but I could demonstrate it here: $\displaystyle \int^{\frac{\pi}{2}}_{0} e^{\ln \left ( \cos x \frac{\mathrm{d}(\cos x)}{\mathrm{d}x}\right ) } \mathrm{d}x \ = \int^{\frac{\pi}{2}}_{0} e^{\ln\left ( \cos x \sin{-x}\right ) } \mathrm{d}x \\ \displaystyle = \int^{\frac{\pi}{2}}_{0} \cos{(-x)} \sin{(-x)} \mathrm{d}x \quad \text{ (Since } \cos(x) = \cos(-x) \text{)}$

$\displaystyle = -\sin^2{(-x)} |_0^{\pi/2} \ = \ -\dfrac{1}{2}$ .

Akeel Howell - 4 years ago

@Akeel Howell – Be careful when simplifying. Recall that $e ^ { \ln x } = x$ only when $x > 0$ (for real logarithms). That is the point of Utsav's solution comment of "the logarithm will not be defined in this interval"

If we used complex logarithms, then your argument might be valid (but would still require checking that we're on the correct logarithmic branch, which isn't immediately obvious to me).

Calvin Lin Staff - 4 years ago

Lu Chee Ket
Oct 30, 2015

Because of "if" we ought to change from -1/ 2 into "None of these" as mathematical error occurs. However, this is not a general practice as the answer suppose to be -1/ 2 when we posses a more complete knowledge about these.

Logarithms & Integrals

This question is derived from the Logarithm Definite Integral question.

Check my problem sets for many more interesting problems.

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