Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Evaluate

$$\int_{0}^{\infty} \frac{\cos x - e^{-x^2}}{x} \ dx$$

share|improve this question
    
Maple says $-\gamma/2$. –  AD. Oct 13 '12 at 12:30
4  
@Chris'ssister: I like your question here. They are really like simple or difficult puzzles. Thanks for sharing them here. –  Babak S. Oct 13 '12 at 13:11
    
@Babak Sorouh: I'm really glad to read these words. Thank you! –  Chris's sis Oct 13 '12 at 13:14

2 Answers 2

up vote 17 down vote accepted

Related problems: (I), (II). Recalling the Mellin transform of a function $f$

$$ F(s) = \int_{0}^{\infty} x^{s-1}f(x) \,dx .$$

Then we consider the more general integral

$$ F(s) = \int_{0}^{\infty} x^{s-1}\left(\cos x - e^{-x^2}\right) \, dx \,. $$

The value of the integral in our problem follows by taking the limit as $s\to 0 $ in the above integral. Evaluating the above integral gives

$$ F(s) = \Gamma \left( s \right) \cos \left( \frac{\pi \,s}{2} \right) - \frac{1}{2}\, \Gamma \left(\frac{s}{2} \right) \,.$$

Taking the limit as $s \to 0 \,,$ we get the desired result

$$ \int_{0}^{\infty} \frac{\cos x - e^{-x^2}}{x} \ dx = -\frac{\gamma}{2}\,. $$

share|improve this answer
    
your way seems so simple! Thanks! (+1) –  Chris's sis Oct 13 '12 at 19:30
    
@Chris'ssister: You are welcome. –  Mhenni Benghorbal Oct 13 '12 at 19:43

The result is

$$ \int_{0}^{\infty} \frac{\cos x - e^{-x^2}}{x} \, dx = -\frac{\gamma}{2},$$

where $\gamma$ is the Euler-Mascheroni constant.

Some direct calculations are available, but I prefer to consider it as a difference of some sort of log-singularities. you can find a slightly general method in this line of approach to calculate integrals of this form in my blog posting.

share|improve this answer
    
thanks for the answer (+1) –  Chris's sis Oct 13 '12 at 12:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.