0
$\begingroup$

Exercise Compute integral $$\int_0^{\infty} \frac{e^{-{(3x)}^2}- e^{-{(6x)}^2}}{x} dx$$

My attempt and questions:

So we can write it as $$\frac{e^{-{(tx)}^2}}{x} |_6^3 \Rightarrow \int_0^{\infty}dx \int_6^3 -2txe^{-(tx)^2}dt$$

We can change order of intigration By Funini's theorem if function is positive which clearly is not as we have minus in front or if we show that $$\int\int_{[0,{\infty}]x[6,3] }2txe^{-(tx)^2} = \int_0^{\infty}dx \int_6^3 |-2txe^{-(tx)^2}| dt = \int_0^{\infty}dx \int_6^3 2txe^{-(tx)^2} dt < \infty$$

$$2 \int_0^{\infty} x dx \int_6^3 te^{-(tx)^2} dt =^{per partes} ln{\frac{1}{2}} < \infty$$

My question Did i calculate corrrectly?

Solving further Hence by Fubini's theore, we can change order of intigration:

$$\int_6^3 -2t dt \int_0^{\infty} xe^{-(tx)^2}dx$$

And here i am stuck because i get infinity by solving by the help of pwr partes.

My questionCould somone please help to proceed?

$\endgroup$
2
  • 1
    $\begingroup$ $\log(6^2/3^2)=2\log(2)$ $\endgroup$ – Mark Viola Jun 25 '18 at 16:56
  • $\begingroup$ Actually, here is a proof with Fubini's theorem in use $\endgroup$ – Jakobian Jun 25 '18 at 16:59
1
$\begingroup$

Note that we have

$$\begin{align} \int_0^\infty \frac{e^{-9x^2}-e^{-36x^2}}{x}\,dx&=2\int_0^\infty\int_3^6 ye^{-xy^2} \,dy\,dx\\\\ &=2\int_3^6\int_0^\infty ye^{-xy^2} \,dx\,dy\\\\ &=2 \int_3^6 \frac1y\,dy \\\\ &=2\log(2) \end{align}$$

$\endgroup$
2
  • $\begingroup$ Hi. Would you please let me know how I can improve my answer? I really want to give you the best answer I can. If this was not useful, I am happy to delete it. Looking forward to your reply. Thank you in advance. $\endgroup$ – Mark Viola Nov 17 '18 at 18:36
  • $\begingroup$ And feel free to up vote and accept an answer as you see fit of course. ;-) $\endgroup$ – Mark Viola Nov 17 '18 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy