2
$\begingroup$

Is there a transformation that makes $\frac{1}{t}(e^{t^2}-1)e^{\frac{1}{t}}(1-\frac{1}{t})$ equal to $t^{-3}(e^{t^2}-e^t)+t^{-2}-t^{-1}$? The reason I ask is that for the integral $$\int_1^t{\int_0^t{\frac{e^{(tx)/y}}{y^{3}}dx}dy}$$, I get the former while Apostol gets the latter (Apostol Calculus Vol 2, 1st Edition, Section 2.9, #6). There graphs look different, so I am guessing not. Still I feel like I solved this problem, but I'd like to know how Apostol got his answer in the form he presented.

First the inner integral:

$$y^{-3}e^{1/y}\int_0^t{e^{tx}dx}$$

y^(-3)e^(1/y)1/t[e^(tx)]_x=0^x=t

y^(-3)e^(1/y)1/t(e^t^2-1)

Now for the outer integral:

1/t(e^t^2-1)*Integrate[y^(-3)e^(1/y),{y,1,t}]

u-substitution:

u=1/y, du=-1/y^2

-Integrate[e^u*u,u]

IBP:

f=u, dg=e^u df=du, g=e^u

-(u*e^u-Integral[e^u,u])=e^u-u*e^u

Plug u back in:

e^(1/y)(1-1/y)|_1^t

Combine it all for final result:

1/t(e^t^2-1)e^(1/t)(1-1/t)

$\endgroup$
  • $\begingroup$ What page? Also, why not show your work? Then, one might be able to tell if you erred or not. $\endgroup$ – Pedro Tamaroff Jul 11 '13 at 17:31
  • $\begingroup$ Hi Peter, I will show my work. Just terrible at LaTeX here so it takes time to enter. I will enter in ascii and then translate it to LaTeX (so that I can get some feedback). $\endgroup$ – Joe Jul 11 '13 at 17:36
  • $\begingroup$ Peter, related question: are you able to input LaTeX very quickly without thinking much about it? I find it so laborious, any tips to speed things up? I use a tablet to do all my math with a digitizer pen. $\endgroup$ – Joe Jul 11 '13 at 17:58
  • $\begingroup$ Oh, using a table is very nice. I'd love that. I have two options: if the coding is not too hard or long, I just type it out (one gets used to it, and learns the codes and stuff) but if it is too long to type, I use MathType. $\endgroup$ – Pedro Tamaroff Jul 11 '13 at 17:59
  • $\begingroup$ Careful $e^{xt/y}\neq e^{1/y}e^{xt}$!!! $\endgroup$ – Pedro Tamaroff Jul 11 '13 at 18:00
2
$\begingroup$

$$\displaylines{ \int_1^t {\frac{1}{{{y^3}}}\left( {\int_0^t {{e^{tx/y}}dx} } \right)dy} = \frac{1}{t}\int_1^t {\frac{1}{{{y^3}}}\left. {\left( {\frac{y}{t}{e^{tx/y}}} \right)} \right|_0^tdy} \cr = \frac{1}{t}\int_1^t {\frac{1}{{{y^3}}}\left( {\frac{y}{t}\left( {{e^{{t^2}/y}} - 1} \right)} \right)dy} \cr = \frac{1}{{{t^2}}}\int_1^t {\frac{{{e^{{t^2}/y}} - 1}}{{{y^2}}}dy} \cr} $$

Can you finish it? Note that $(y^{-1})'=-y^{-2}$

$\endgroup$

Your Answer

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

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