Help me prove $\left(\int_{0}^{\infty} t^{50} e^{-t} \,\mathrm dt \right)^{1/2}$ isn't a perfect square

Question

I need some help in proving

$$\left( \displaystyle\int_{0}^{\infty} t^{50} e^{-t} \,\mathrm dt \right)^{1/2}$$

isn't a perfect square. The only way I can think is repeated integration by parts which is obviously impractical and I still likely wouldn't be able to deduce if the result was a perfect square.

Thanks very much guys! :D

Answer 1

Repeated integration by parts seems to be the way to do it. There will be an obvious pattern, so you don't need to do it 50 times. Hint: The integral has the value 50!.

Answer 2

Hint: Compute $\displaystyle \int_0^\infty t^n e^{-t}\,\mathrm dt$ by induction on $n$ -- you should be able to discover a nice pattern, which allows you to plug in the value $n = 50$.

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After you do that, also check out the Gamma function.

I trust you can complete the proof that the resulting quantity isn't a perfect square.

Answer 3

Consider $f(t) = \displaystyle-\left[\sum_{n=0}^{50} \frac{50!(-t)^n}{n!}\right] e^{-t}$, then $$f^\prime(t) = -\left[\sum_{i=1}^{50} \frac{50!(-t)^{n-1}}{(n-1)!}\right]e^{-t} +\left[\sum_{n=0}^{50} \frac{50!(-t)^n}{n!}\right] e^{-t} = t^{50}e^{-t}.$$

Thus, $\displaystyle\int_0^\infty t^{50}e^{-t}\,\mathrm{d}t = \left.-\left[\sum_{n=0}^{50} \frac{50!(-t)^n}{n!}\right] e^{-t}\right|_0^\infty = 50!$.

Since $50!$ only contains one factor of the prime $47$, it is not a perfect square. Hence $\displaystyle\left(\int_0^\infty t^{50}e^{-t}\,\mathrm{d}t\right)^\frac{1}{2}$ is not even an integer.

Answer 4

In 1975 P. Erdős and J. L. Selfridge proved "The product of consecutive integers is never a power" in Illinois J. Math. Volume 19, Issue 2 (1975), 292-301.

One URL: http://projecteuclid.org/euclid.ijm/1256050816