# Is there a transformation that makes $\frac{1}{t}(e^{t^2}-1)e^\frac{1}{t}(1-\frac{1}{t})$ equal to $t^{-3}(e^{t^2}-e^t)+t^{-2}-t^{-1}$?

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)

• What page? Also, why not show your work? Then, one might be able to tell if you erred or not. – Pedro Tamaroff Jul 11 '13 at 17:31
• 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). – Joe Jul 11 '13 at 17:36
• 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. – Joe Jul 11 '13 at 17:58
• 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. – Pedro Tamaroff Jul 11 '13 at 17:59
• Careful $e^{xt/y}\neq e^{1/y}e^{xt}$!!! – Pedro Tamaroff Jul 11 '13 at 18:00

$$\displaylines{ \int_1^t {\frac{1}{{{y^3}}}\left( {\int_0^t {{e^{tx/y}}dx} } \right)dy} = \frac{1}{t}\int_1^t {\frac{1}{{{y^3}}}\left. {\left( {\frac{y}{t}{e^{tx/y}}} \right)} \right|_0^tdy} \cr = \frac{1}{t}\int_1^t {\frac{1}{{{y^3}}}\left( {\frac{y}{t}\left( {{e^{{t^2}/y}} - 1} \right)} \right)dy} \cr = \frac{1}{{{t^2}}}\int_1^t {\frac{{{e^{{t^2}/y}} - 1}}{{{y^2}}}dy} \cr}$$
Can you finish it? Note that $(y^{-1})'=-y^{-2}$