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.

Determine if the integral converges:

$$\int_{-\infty}^{2} \frac{e^{3x}dx}{1+x^2}$$

I've tried this:

$$f(x)=\frac{e^{3x}}{1+x^2}>\frac{e^{\ln{3x}}}{1+x^2}=\frac{3x}{1+x^2}=g(x)$$

now since g(x) is similar to $\frac1x$ the integral diverges. [Sparing the limits prof].

Anything i did wrong? Feels to me like i didn't need to pay attetion to $-\infty$, and i'm probably missing something.

share|improve this question
1  
Your problem is that the integral of $g$ goes to minus infinity and not plus infinity. –  N.U. Apr 11 '13 at 18:19

1 Answer 1

up vote 4 down vote accepted

The integral converges. To see this, do a substitution $x=-y$, and then compare with

$$\int_2^{\infty} dy \: e^{-3 y}$$

share|improve this answer
    
felt like i'm missing exactly that, Didn't think about substitution. Thanks ! –  StationaryTraveller Apr 11 '13 at 18:08

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.