I plugged rather basic integral into my new calculator (TI Nspire CX CAS) just to see what it could do. Surprisingly, it returned undef and I'm wondering if somebody could explain why and/or how I can fix this.

The integral I used is as follows

$$\int_0^\infty \frac{e^{\frac{-x}{y}}e^{-y}}{y}dx = e^{-y}$$ The calculator couldn't perform this, but when I replaced $y$ by a constant like the number 2 it was able to handle it, which I find strange. I presume it's assuming that $y$ can take a value of zero which would make it undefined.


Even after simplifying the integral and pulling constants out, it still cannot perform the integration

$$\frac{e^{-y}}{y}\int_0^\infty e^{\frac{-x}{y}}dx = e^{-y}$$


And what happens if $y = 0$? How is the calculator supposed to know what you mean for such a value?

Even if you defined the function $f(x,y) = e^{-x/y}e^{-y}/y$ in such a way as to make it meaningful at $y = 0$, there remains a fundamental misconception in your claim, easily tested when $y < 0$. What happens then?

  • $\begingroup$ that's what I presume the problem is too. Is there any way to define $y \neq 0$ on the Nspire CX? $\endgroup$ – audiFanatic May 15 '14 at 4:15
  • $\begingroup$ Some implementations of computer algebra systems are not quite as advanced or complete as others. For instance, if you type Integrate[Exp[-x/y], {x, 0, Infinity}] in Mathematica, it gives ConditionalExpression[y, Re[1/y] > 0]. That's a big clue into what is really happening. $\endgroup$ – heropup May 15 '14 at 4:18
  • $\begingroup$ exactly, the same thing happens in Wolfram Alpha (aka web-based Mathematica) $\endgroup$ – audiFanatic May 15 '14 at 4:20
  • $\begingroup$ Well, then that's your answer. The CAS on the TI calculator isn't sophisticated enough to discover convergence conditions on its own. To be fair, you can't honestly expect it to. Early versions of Mathematica didn't, either. The question I have for you is, do you understand why the integral behaves this way? $\endgroup$ – heropup May 15 '14 at 4:26
  • $\begingroup$ yes, I do understand it. $\endgroup$ – audiFanatic May 15 '14 at 5:20

I had this same problem.

Take the integral from 0 to z and.

Use the limit function to take the limit of the result as z approaches infinity from the minus side.

You can do all of this on one line, if you have the integral within the parenthesis for the limit function.


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