MHB Calculus II Integral Convergence and Divergence Questions

[ardentmed]

For a,b, and c respectively, I got divergent (to -infinity), convergent (to π/6), and divergent (to infinity, since the first part's sum is 1/3, but lim negative infinity gives infinity, thus the summation of the two integrals gives a divergent integral). I'm sure these are right, but I'd appreciate some help, especially for 8c.I really appreciate the help guys. Thanks in advance.

 

[Prove It]

Finally, for 8abc repectively, I got divergent (to -infinity), convergent (to π/6), and divergent (to infinity, since the first part's sum is 1/3, but lim negative infinity gives infinity, thus the summation of the two integrals gives a divergent integral). I'm sure these are right, but I'd appreciate some help, especially for 8c.I really appreciate the help guys. Thanks in advance.

I agree that the first one needs to be divergent, as the integral of a 1/x type function is a logarithm, which grows without bound (but at the slowest possible rate).

For the second:

$\displaystyle \begin{align*} \int{ \frac{\mathrm{e}^x\,\mathrm{d}x}{\mathrm{e}^{2x} + 3}} &= \int{\frac{\mathrm{e}^x\,\mathrm{d}x}{ \left( \mathrm{e}^x \right) ^2 + 3 } } \end{align*}$

Now substitute $\displaystyle \begin{align*} \mathrm{e}^x = \sqrt{3}\tan{ \left( \theta \right) } \implies \mathrm{e}^x\,\mathrm{d}x = \sqrt{3}\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{\mathrm{e}^x\,\mathrm{d}x}{ \left( \mathrm{e}^x \right) ^2 + 3 }} &= \int{ \frac{\sqrt{3}\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta}{ \left[ \sqrt{3}\tan{\left( \theta \right) } \right] ^2 + 3 }} \\ &= \frac{\sqrt{3}}{3} \int{ \frac{\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta}{ \tan^2{\left( \theta \right) } + 1 }} \\ &= \frac{\sqrt{3}}{3} \int{ \frac{\sec^2{\left( \theta \right) } \,\mathrm{d}\theta}{\sec^2{\left( \theta \right) } }} \\ &= \frac{\sqrt{3}}{3} \int{ 1 \, \mathrm{d}\theta} \\ &= \frac{\sqrt{3}}{3} \theta + C \\ &= \frac{\sqrt{3}}{3} \arctan{ \left( \frac{\mathrm{e}^x}{\sqrt{3}} \right) } + C \end{align*}$

so your definite integral is

$\displaystyle \begin{align*} \int_0^{\infty}{\frac{\mathrm{e}^x\,\mathrm{d}x}{\mathrm{e}^{2x} + 3}} &= \lim_{\epsilon \to \infty} \int_0^{\epsilon}{ \frac{\mathrm{e}^x\,\mathrm{d}x}{\mathrm{e}^{2x} +3} } \\ &= \frac{\sqrt{3}}{3} \lim_{\epsilon \to \infty} \left[ \arctan{ \left( \frac{\mathrm{e}^x}{\sqrt{3}} \right) } \right]_0^{\epsilon} \\ &= \frac{\sqrt{3}}{3} \lim_{\epsilon \to \infty} \left[ \arctan{ \left( \frac{\mathrm{e}^{\epsilon}}{\sqrt{3}} \right) } - \arctan{ \left( \frac{\mathrm{e}^0}{\sqrt{3}} \right) } \right] \\ &= \frac{\sqrt{3}}{3} \left( \frac{\pi}{2} - \frac{\pi}{6} \right) \\ &= \frac{\sqrt{3}}{3} \left( \frac{\pi}{3} \right) \\ &= \frac{\sqrt{3}\,\pi}{9} \end{align*}$

so this integral is convergent :)

 

[ardentmed]

Any idea about the third one? I'm almost definite that it is divergent, but one of the sums for the split up integral gives me 1/3, a convergent value.

 

[Prove It]

Any idea about the third one? I'm almost definite that it is divergent, but one of the sums for the split up integral gives me 1/3, a convergent value.

You'll have to post all your working out, I have no idea how to tell you where (or if) you have gone wrong if I have no idea what you have done...

 

[ardentmed]

You'll have to post all your working out, I have no idea how to tell you where (or if) you have gone wrong if I have no idea what you have done...

Well, if you take u=-x^3 and -(du/3)=x^2 dx, then the resulting integral is (-1/(3e^ (x^3))

Therefore, from infinity to zero, the sum is 1/3, and from zero to negative infinity, the sub is just infinity since [(-e)^ -(-infinity)] / 3 is infinity.

Thanks.

 

[Prove It]

Well, if you take u=-x^3 and -(du/3)=x^2 dx, then the resulting integral is (-1/(3e^ (x^3))

Therefore, from infinity to zero, the sum is 1/3, and from zero to negative infinity, the sub is just infinity since [(-e)^ -(-infinity)] / 3 is infinity.

Thanks.

So if you end up with an infinite amount, what conclusion should you come to?

 

[ardentmed]

So if you end up with an infinite amount, what conclusion should you come to?

Then it would obviously be divergent. My question is if my rationale was correct, meaning was my work leading up to the conclusion that it equals infinity correct?

Thanks.