Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to calculate the following limit:

$$\displaystyle{\lim_{x \to 0} \cfrac{\displaystyle{\int_1^{x^2+1} \cfrac{e^{-t}}{t} \; dt}}{3x^2}}$$

For that, I use L'Hopital and the Fundamental Theorem of Calculus, obtaining the following:

$$\displaystyle{\lim_{x \to 0} \cfrac{\displaystyle{\int_1^{x^2+1} \cfrac{e^{-t}}{t} \; dt}}{3x^2}}=\displaystyle{\lim_{x \to 0} \cfrac{\frac{e^{-(x^2+1)}}{x^2+1} \cdot 2x}{6x}}=\lim_{x \to 0} \cfrac{e^{-(x^2+1)}}{3(x^2+1)}=\cfrac{e^{-1}}{3}$$

But if I calculate the limit in Wolfram Alpha, I get the following. Limit Wrong?

I calculated the limit also in Mathematica 8.0, and the result is still the same: $\frac 13(\frac 1e-1) $ So, what is my mistake calculating the limit?

share|improve this question
Pro tip: report the error to Wolfram and you'll get a free t-shirt! (Happened to an office mate of mine last year) –  icurays1 Jan 15 '13 at 17:04
Mathematica's own numerical calculations agree with you: wolframalpha.com/input/?i=Integrate[1%2Ft*E^%28-t%29%2C+{t%2C+1%2C+1%‌​2B10^-10}]%2F%283*10^-10%29 –  user7530 Jan 15 '13 at 17:04
By wrapping the expression in Assuming[x ∊ Reals, ... ] in Mathematica I do indeed get the expected result. –  Frxstrem Jan 15 '13 at 17:06
there not too far from each other. Alternate form of the first is 1/3e and the second is 1/3e -1/3. Though i cant see any reason why the "-1/3" would come in –  exussum Jan 15 '13 at 18:05
Wouldn't this be better on Mathematica.SE? –  robjohn Jan 17 '13 at 2:21

4 Answers 4

Some more observations on Mathematica's behavior: it yields

$$ \frac{d}{dx} \int_1^{x} \frac{e^{-t}}{t} \, dt = \frac{e^{-x} - 1}{x} $$

restricted to $\Im(x) \neq 0$ or $\Re(x) \geq 0$. But changing things slightly gives

$$ \frac{d}{dt} \int \frac{e^{-t}}{t} \, dt = \frac{e^{-t}}{t}. $$

I had originally suspected there was something fishy with the branch cut: Mathematica computes

$$ \int_1^x \frac{e^{-t}}{t} \, dt = -\mathrm{Ei}(-1) - \log(x) - \Gamma(0,x) $$

again restricted to $\Im(x) \neq 0 \vee \Re(x) \geq 0$. However:

  • The point we are interested in is away from the branch discontinuity
  • I would have expected it to get the derivative right even if there were weird branch issues

(using $x+1$ in the above instead of $x$ does not make any qualitative difference)

Without limits, it computes

$$ \int \frac{e^{-t}}{t} \, dt = \mathrm{Ei}(-t) \color{gray}{+ \mathrm{constant}}$$

If you shift the integrand, you get

$$ \frac{d}{dx} \int_0^x \frac{e^{-(u+1)}}{u+1} \, du = \frac{e^{-(u+1)}}{u+1} $$

and correspondingly

$$ \lim_{y \to 0} \frac{ \int_0^y \frac{e^{-(u+1)}}{u+1} \, du }{3 y} = \frac{1}{3e} $$

(I substituted $x^2 = y$ so that wolfram would finish the calculation for me. This substitution does not make a qualitative difference in the original)

I think the key difference is that in the first version, the branch point is $t=0$, and Mathematica focuses on the behavior there -- which is inherently weird and strange because it's a branch point (and given that, I'm not sure if using the result leads to computing something correct but strange, or something ill-defined). But in the second version, the branch point is at $u=-1$, but Mathematica still focuses on $u=0$ so it gets sane results.

share|improve this answer

Mathematica is wrong because everything in sight is positive.

Unfortunately I cannot say what Mathematica is doing wrong.

share|improve this answer
+1 I did it in Maple 16, and it gave me $e^{-1}/3$. –  B. S. Jan 15 '13 at 17:02
That hurts: I hate (teaching) Maple!! –  Jp McCarthy Jan 15 '13 at 17:03
Poor Maple. Without it I will be lost. :D –  B. S. Jan 15 '13 at 17:04
Maple's what they use at the community college I attend... What's so bad about it? –  anorton Jan 15 '13 at 17:15
@anorton Actually nothing imo. I like it. –  MyUserIsThis Jan 15 '13 at 17:33

For a simple check of your answer, note that the integrand at $t=1$ is $\frac 1e$ , continuous and slowly varying. The integral is very close to $\frac {x^2}e$, supporting your answer.

share|improve this answer

Limit[Integrate[Exp[-t]/t, {t, 1, 1 + x^2}, Assumptions -> x \[Element] Reals]/(3 x^2), x -> 0] yields the correct $\frac1{3e}$. Putting the assumptions on the Limit instead of the Integrate returns the same erroneous answer: $\frac13\left(-1+\frac1e\right)$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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