I'm trying to compute: $\int_{0}^{1}e^x(1-x)^{100}dx$. I tried to use integration by parts but it didn't work out for me(since I need to do that 100 times, and obviously there's a shorter solution) , I substituted $(1-x)=u$ and got $e\int_0^1e^{-t}t^{100}$, again I can't do with that much. Any suggestion how should solve this integral?

Thanks a lot guys!

  • $\begingroup$ Try for a Reduction Formula. For that, Parts is (are?) good. $\endgroup$ – André Nicolas Feb 15 '12 at 15:41
  • $\begingroup$ Please expand on "it didn't work out for me"? Which result did you get -- in which way did that not work out? $\endgroup$ – Henning Makholm Feb 15 '12 at 15:41
  • 1
    $\begingroup$ I imagine the easiest thing to do is expand $e^{-t}t^{100}$ into a power series. $\endgroup$ – David Mitra Feb 15 '12 at 15:46
  • 2
    $\begingroup$ Just as a side note: If all you want to do is estimate the integral, there's a nifty trick: $\int_0^1 e^x (1-x)^{100} \approx \int_0^1 e^x e^{-100x} \approx \int_0^{\infty} e^{-99x}=\frac{1}{99}$ that gets you within about 1% of the correct answer. $\endgroup$ – Kevin P. Costello Feb 15 '12 at 17:21
  • $\begingroup$ My answer below is exact and in closed form, if you allow the floor function. $\endgroup$ – alex.jordan Feb 16 '12 at 1:45

For what it's worth:

Write $$ t^{100}e^{-t}=t^{100}(1-t+{t^2\over 2!}-{t^3\over 3!}-\cdots ) =t^{100}-t^{101}+{t^{102}\over 2!}-{t^{103}\over 3!}-\cdots $$

The above series is uniformly convergent on $[0,1]$; thus: $$ \eqalign{ \int_0^1 e^{-t}t^{100}\,dt &=\sum_{n=0}^\infty \int_0^1 (-1)^n{t^{100+n}\over n!}\cr &=\sum_{n=0}^\infty (-1)^n{t^{101+n}\over({101+n}) n!}\Bigl|_0^1\cr &=\sum_{n=0}^\infty (-1)^n{1\over({101+n}) n!}. \cr } $$

  • $\begingroup$ Worth a lot. very nice, Thanks! $\endgroup$ – Jozef Feb 15 '12 at 16:38

The integral is exactly the fractional part of $100!\,e$, or in other words $100!\ e-\lfloor100!\ e\rfloor\approx0.00999901019\ldots$

Apply integration by parts to the integral $I_n=\int_0^1e^{1-t}t^n\,dt$ (it's nicer not to pull the $e$ out to the front) and we find for $n\geq1$, $$I_n=-1+nI_{n-1}$$

This gives us $$I_{100}=-1+100[-1+99[-1+98[-1+\cdots+2[-1+1I_0]\cdots]]]$$

$I_0$ is a straightforward computation: $e-1$. So

This gives us $$I_{100}=-1+100[-1+99[-1+98[-1+\cdots+2[-1+e-1]\cdots]]]$$

Here is a nice observation. Once this is multiplied out, it (clearly?) simplifies to $100!\,e-N$ for some integer $N$. A graphical examination of the integral reveals that $I_{100}$ is somewhere between $0$ and $1$. (You could prove this using the fact that $e^{1-t}t^{100}=e^{1-t}tt^{99}\leq t^{99}$ on $[0,1]$.) So $N$ must equal the integer part of $100!\,e$, leaving $I_{100}$ to be the fractional part.

It's interesting to note that since $I_n\to0$ as $n\to\infty$, the fractional part of $n!\,e$ must approach zero; that is, $n!\,e$ gets closer and closer to being an integer. (Although I suppose that is obvious if we consider the usual series expansion for $e$.)

For computational purposes, we can use this to find a decimal approximation by throwing out the first $100$ terms or so (which are all integers) of the series expansion for $100!\, e$.

$$ \begin{align} \int_0^1e^{1-t}t^{100}\,dt & = 100!\, e-\lfloor100!\,e\rfloor\\ & = \sum_{n=101}^{\infty}\frac{100!}{n!} \end{align} $$

This is the series that bgins has found with a slightly different argument. At first, this series converges faster than David Mitra's alternating series. It is correct to at least 17 decimal places after only 8 partial summands. David's requires 18 partial summands to get that much accuracy. However since both series have a ratio of order $1/n$ and David's series is alternating, I think that in the long run for very high accuracy demands, his series might be better.

  • $\begingroup$ Very good answer! $\endgroup$ – Pedro Tamaroff Feb 16 '12 at 22:14

Use the formula $$ \frac{d}{dx}\left(e^x\ \sum_{n\ge0}\ (-1)^n\ f^{(n)}(x)\right)=e^x\ f(x), $$ which holds if $f$ is a polynomial.

  • $\begingroup$ @WillieWong Is it even true that this series $\sum (-1)^nf^{(n)}(x)$ necessarily converges for all real analytic functions? If it converges, then for any $x$, $\sum (-1)^nf^{(n)}(x)/n!$ converges, so that means the Taylor series around $x$ must have radius of convergence $\geq 1$. But we can easily construct real analytic functions that don't have radius of convergence $1$ at some $x$, can't we? Like $\frac{1}{1+4x^2}$, which has a radius of convergence $\frac{1}{2}$ at $x=0$. $\endgroup$ – Thomas Andrews Feb 15 '12 at 16:39
  • $\begingroup$ Oops. @Thomas and Pierre-Yeves, I was too hasty at computing the decay condition needed. Deleting my previous comment. What I was trying to say is that "which holds if $f$ is a polynomial" can be replaced by a suitable convergence condition, and that the sum doesn't really have to terminate. Sorry! $\endgroup$ – Willie Wong Feb 16 '12 at 8:52

You could use integration by parts. You will end up with a recursion, like this:

$$\int\limits_0^1 {{e^{ - x}}{x^n}dx = \left[ { - {e^{ - x}}{x^n}} \right]_0^1} + n\int\limits_0^1 {{e^{ - x}}} {x^{n - 1}}dx$$ $${I_n} = -\frac{1}{e} + n{I_{n - 1}}$$


$$\int\limits_0^1 {{e^{ - t}}{t^n}dt} = {I_n}$$

Using it sufficient times you'll end up with your result.

  • $\begingroup$ Thanks! sadly solving this exercise it's not up my choice :) $\endgroup$ – Jozef Feb 15 '12 at 16:29
  • $\begingroup$ You should use integration by parts... Actually this is not necessary. // Unless I am mistaken, the value of $I_n$ that you propose is negative, you might want to modify this. $\endgroup$ – Did Feb 15 '12 at 22:54
  • $\begingroup$ @DidierPiau I used $$\int\limits_0^1 {{e^{ - x}}{x^n}dx = \left[ { - {e^{ - x}}{x^n}} \right]_0^1} + \int\limits_0^1 {{e^{ - x}}n} {x^{n - 1}}dx$$ $\endgroup$ – Pedro Tamaroff Feb 15 '12 at 23:08

Substitute $1-x = t$ , so :

$I= -e \int \limits_1^0 e^{-t} \cdot t^{100}\,dt =e \int \limits_0^1 e^{-t} \cdot t^{100}\,dt$

Now substitute : $-t=s$ , so :

$I= -e \int \limits_0^{-1} e^{s} \cdot s^{100}\,ds = e \int \limits_{-1}^{0} e^{s} \cdot s^{100}\,ds$

This integral you can solve using Integration by parts several times .For start choose :

$u = s^{100}~$ , and $dv = e^s ds$


As already said, this is $I_n$ for $n=100$, where $$ I_n=\int_0^1x^n\mathrm e^{1-x}\mathrm dx. $$ To compute $I_n$ for every $n\geqslant0$ at once, one can group these in a series, that is, consider, for every $s$ small enough, $$ \sum_{n\geqslant0}\frac{s^n}{n!}I_n=\int_0^1\sum_{n\geqslant0}\frac{s^n}{n!}x^n\mathrm e^{1-x}\mathrm dx=\int_0^1\mathrm e^{1-x(1-s)}\mathrm dx=\frac{\mathrm e-\mathrm e^{s}}{1-s}. $$ The expansion of the first part of the RHS is $$ \frac{\mathrm e}{1-s}=\sum\limits_{n\geqslant0}\mathrm es^n. $$ As regards the second part, one knows that $$ \mathrm e^s\cdot\frac1{1-s}=\left(\sum\limits_{n\geqslant0}\frac{s^n}{n!}\right)\cdot\left(\sum\limits_{n\geqslant0}s^n\right), $$ whose coefficient of $s^n$ is $$ \sum\limits_{k=0}^n\frac1{k!}. $$ Putting all these together yields $$ \frac{I_n}{n!}=\mathrm e-\sum\limits_{k=0}^n\frac1{k!}=\sum\limits_{k=0}^{+\infty}\frac1{k!}-\sum\limits_{k=0}^n\frac1{k!}=\sum\limits_{k\geqslant n+1}\frac1{k!}, $$ and finally, $$ I_n=\sum\limits_{k\geqslant1}\frac{n!}{(n+k)!}, $$ To get an approximate value of $I_n$, one can proceed as follows. Keeping only the first term of the series in the RHS yields a lower bound. Replacing the $k$th term by $\frac1{(n+1)^k}$ and summing the resulting series yields an upper bound. These read $$ \frac1{n+1}\lt I_n\lt\frac1n. $$

  • $\begingroup$ And more generally, $$ \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \ldots + \frac{1}{(n+1)(n+2)\ldots(n+k)} < I_n < \frac{1}{n+1}+\frac{1}{(n+1)(n+2)} + \ldots + \frac{1}{(n+1)(n+2)\ldots(n+k-1)} + \frac{1}{(n+1)(n+2)\ldots(n+k-2)(n+k-1)^2}$$ $\endgroup$ – Robert Israel Feb 17 '12 at 1:06

Define $$ I_n = e\int_0^1 e^{-t} t^n dt $$

We have (integration by part) $I_n=n I_{n-1} - 1$, and $I_0 = e-1$.

Write $I_n$ as $a_n e - b_n$, with $a_n$ and $b_n$ rational numbers. We have $a_n = n a_{n-1}$ and $b_n = n b_{n-1} +1$, with $a_0=1=b_0=1$.

So obviously, $a_n = n!$. Way less obviously, we have $b_n = A000522(n)$. See the [OIES]1 page for generating series and other forms for $b_n$. The recurrence in itself is a very nice form, it allows the use of fast algorithms to compute a lot of terms. It also allows for asymptotic analysis.

So in the end, $$ e\int_0^1 e^{-t} t^n dt = n! e - A000522(n) $$

Computer easily find that $b_{100} = A000522(100)$ is 2536869555601272974152707482122802204451475785662981422327751859874492\ 5390838644651894048542515204979326740773232800349360951349984969417670\ 9764490323163992001

Bonus : the factorization of $b_{100}$ :

{{59, 1}, {197, 1}, {281, 1}, {617, 1}, {6791290111, 1}, {29565843698156503, 1}, {42933474506607537350507, 1}, {146032600411218505211021315344688241113203163378586255678583083\ 99982138258246286181319177385766944701, 1}}


With $t=1-x,~dt=-dx$, the integral $$ I=\int_{0}^{1}e^x(1-x)^{100}dx=e\int_{0}^{1}t^{100}e^{-t}dt=e\;I_{100} $$ can be expressed in terms of $I_n=\int_{0}^{1}t^{n}e^{-t}dt$. Using integration by parts with $u=t^n,~du=nt^{n-1}dt$ and $v=-e^{-t},~dv=e^{-t}dt$, we find for $n>0$ $$ I_n =\int_{0}^{1}t^{n}e^{-t}dt =\left[-t^ne^{-t}\right]_0^1 +n\int_{0}^{1}t^{n-1}e^{-t}dt =-\frac1e+nI_{n-1} $$ while $I_0=1-\frac1e$, since at the lower endpoint, $t^0=1$ does not vanish. Unraveling the recursion, we find $$ I_n=n!\left(1-\frac1e\;\sum_{k=0}^n\frac1{k!}\right) =\frac{n!}{e}\left(e-\sum_{k=0}^n\frac1{k!}\right) $$ $$ =\frac{n!}{e}\sum_{k=n+1}^\infty\frac1{k!} %=\frac1e\sum_{k=n+1}^\infty\frac{n!}{k!}% (hidden) extra step! =\frac1e\sum_{k=1}^\infty\frac{n!}{(n+k)!} $$ which can be bounded thus: $$ \frac{e^{-1}}{n+1} <I_n <\frac1e\sum_{k=1}^\infty\frac1{n^k} =\frac{e^{-1}}{n-1} $$ In particular, $$ I=e\;I_{100} =\sum_{k=1}^\infty\frac{100!}{(100+k)!} =\frac1{101} +\frac1{101\cdot102} +\frac1{101\cdot102\cdot103} +\cdots $$ Numerically, this agrees with David Mitra's more elegant solution,

$$ \eqalign{ I_n &=\sum_{k=0}^\infty {(-1)^k\over k!} \int_0^1 t^{n+k} dt \cr &=\sum_{k=0}^\infty {(-1)^k\over (n+k+1)k!}. \cr } $$ I offer only some numerical evidence of this from Sage (which used maxima for at least the third quantity).


0.009999010191094737330783479071750558784854694330\ 66671978145864533161714326816356034370681432918468\ 53285140817781582051413229186884063403403557977080\ 02120596655485207781703353097131323526778542109794\ 17015925291075182530361953938643022940937197666661\ 05165157555781181237126456446981437647360344459677\ 36642319892827301897254303584248930056503030250811\ 27060937927312963829104793337147051891405375697835

k = var('k')
I1 = sum(factorial(100)/factorial(100+k), k, 1, infinity)
I1.n(digits=396) # sum of positive terms

0.009999010191094737330783479071750558784854694330\ 66671978145864533161714326816356034370681432918468\ 53285140817781582051413229186884063403403557977080\ 02120596655485207781703353097131323526778542109794\ 17015925291075182530361953938643022940937197666661\ 05165157555781181237126456446981437647360344459677\ 36642319892827301897254303584248930056503030250811\ 27060937927312963829104793337147051891405375697835

# sum is to 1000 because summing to infinity took too long
I2=e*sum((-1)^k/factorial(k)/(k+101), k, 0, 1000); I2.n(digits=396)
I2.n(digits=396) # alternating series

0.009999010191094737330783479071750558784854694330\ 66671978145864533161714326816356034370681432918468\ 53285140817781582051413229186884063403403557977080\ 02120596655485207781703353097131323526778542109794\ 17015925291075182530361953938643022940938823909717\ 30629190237480312082277049371942004495735825187534\ 58710947176445868284110586963289380133420798065473\ 61761071441791460666209221052877877060247573885386

We can see that the third quantity, the more elegantly derived alternating series, as it was calculated above, differs from the first two (as they were calculated) by about $0.17\times10^{-238}$, i.e., its fifth row ends with $8823909717$ rather than $7197666661$.

  • $\begingroup$ Shouldn't $(e-1)100!\sum_{k=0}^{100}\frac1{k!}\approx e(e-1)\cdot 100!$? $\endgroup$ – Thomas Andrews Feb 15 '12 at 16:24

For $n>-1$, consider $$I=\int\limits_{0}^{1}{\rm d}t\,e^{t}(1-t)^{n}.$$ With the substitution $u=1-t$ $$I=e \int\limits_{0}^{1}{\rm d}u{\hspace{1pt}} e^{-u}u^{n}=e \Bigg(\int\limits_{0}^{\infty}{\rm d}u{\hspace{1pt}} e^{-u}u^{n}-\int\limits_{1}^{\infty}{\rm d}u{\hspace{1pt}} e^{-u}u^{n}\Bigg)=e\big[\Gamma(n+1)-\Gamma(n+1,1)\big],$$ where $\Gamma(s)$ and $\Gamma(s,a)$ are the gamma and incomplete gamma functions respectively. For $n=100$ we have $e\big[\Gamma(101)-\Gamma(101,1)\big]\approx 0.009999$.

  • $\begingroup$ If $u=1-x$ then how does $e^t$ become $e^{-u}$? Could you have intended $e^x$ and $dx$ where you wrote $e^t$ and $dt$? $\endgroup$ – Michael Hardy Feb 16 '12 at 1:57
  • $\begingroup$ Thanks for catching that. I fixed it. $\endgroup$ – Tensor Feb 16 '12 at 2:22
  • $\begingroup$ This is also explained following wolframalpha.com/input/?i=integrate+x%5Ek%2Fexp(x) - nice solution. $\endgroup$ – Bastian Ebeling Jun 18 '12 at 9:40

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.