How to show $\int_{0}^{\infty} x^{\sqrt{10}}e^{-x^{\frac{1}{100}}} dx < \infty$?

Question

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1. What the problem asks

We have the improper integral

[ I=\int_{0}^{\infty} x^{\sqrt{10}}\;e^{-\,x^{1/100}}\;dx ]

and we must prove that it is finite (i.e. convergent).
The only possible trouble points are

(x\to0^{+}) (the lower limit), and
(x\to\infty) (the upper limit).

We will treat the two ends separately.

2. Behaviour at the lower limit (x=0)

When (x) is very small, (x^{1/100}) is also small, so

[ e^{-x^{1/100}} = 1 - x^{1/100}+O!\left(x^{2/100}\right) \longrightarrow 1 . ]

Hence near (0)

[ x^{\sqrt{10}}e^{-x^{1/100}} \sim x^{\sqrt{10}} . ]

The integral (\int_{0}^{1} x^{\sqrt{10}}dx) converges because the exponent (\sqrt{10}>-1). Formally,

[ \int_{0}^{1} x^{\sqrt{10}}e^{-x^{1/100}}dx\le \int_{0}^{1} x^{\sqrt{10}}dx=\frac{1}{\sqrt{10}+1}<\infty . ]

Thus the integral is proper at the lower limit.

3. Behaviour at the upper limit (x\to\infty)

The factor (e^{-x^{1/100}}) decays much faster than any power of (x). A convenient way to see the finiteness is to make the substitution

[ t = x^{1/100}\qquad\Longrightarrow\qquad x = t^{100}, \quad dx = 100\,t^{99}\,dt . ]

Then

[ \begin{aligned} I &=\int_{0}^{\infty} \bigl(t^{100}\bigr)^{\sqrt{10}}\, e^{-t}\; 100\,t^{99}\,dt
&= 100\int_{0}^{\infty} t^{100\sqrt{10}+99}\,e^{-t}\,dt . \end{aligned} ]

The integral on the right‑hand side is the Gamma function:

[ \int_{0}^{\infty} t^{\alpha-1}e^{-t}\,dt = \Gamma(\alpha),\qquad \alpha>0 . ]

Here

[ \alpha = 100\sqrt{10}+100 \; (>0), ]

so

[ I = 100\,\Gamma!\bigl(100\sqrt{10}+100\bigr) . ]

Since the Gamma function is finite for every positive argument, (I) is finite. Hence the integral converges at the upper limit.

4. A direct comparison (optional)

If one prefers a comparison test instead of the substitution, note that for any (a>0) there exists (M>0) such that

[ x^{a}\le e^{\frac12 x^{1/100}}\qquad\text{for all }x\ge M . ]

Multiplying by (e^{-x^{1/100}}) gives

[ x^{a}e^{-x^{1/100}}\le e^{-\frac12 x^{1/100}}\qquad(x\ge M). ]

The right‑hand side is integrable on ([M,\infty)) because

[ \int_{M}^{\infty} e^{-\frac12 x^{1/100}}dx = 100\int_{M^{1/100}}^{\infty} t^{99}e^{-\frac12 t}\,dt<\infty . ]

Thus the original integrand is dominated by an integrable function, confirming convergence.

5. Final answer

[ \boxed{\displaystyle \int_{0}^{\infty} x^{\sqrt{10}}e^{-x^{1/100}}dx = 100\,\Gamma!\bigl(100\sqrt{10}+100\bigr) <\infty } ]

The integral converges because it behaves like a power function at the origin and, after the change of variables (t=x^{1/100}), becomes a Gamma integral, which is known to be finite for positive parameters.

6. Common mistakes

Mistake	Why it is wrong	How to avoid it
Ignoring the lower limit and assuming convergence only depends on the behaviour at (\infty).	Near (0) the integrand behaves like (x^{\sqrt{10}}); if the exponent were (\le -1) the integral would diverge.	Check both endpoints separately.
Applying the limit‑comparison test with the wrong comparison function, e.g. comparing with (1/x).	(x^{\sqrt{10}}e^{-x^{1/100}}) decays much faster than any (1/x^p); the limit of their ratio may be (0) and give no information.	Use a comparison that dominates the integrand, such as (e^{-c x^{1/100}}) with (c>0).
Forgetting the Jacobian when substituting (t = x^{1/100}).	Omitting the factor (dx = 100 t^{99} dt) would give an incorrect power of (t).	Write the substitution step explicitly and keep track of the differential.
Assuming the Gamma function diverges because its argument looks large.	(\Gamma(z)) is finite for all (z>0); it only diverges at non‑positive integers.	Remember the definition (\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}dt) and that it is finite for positive (z).

Keeping these points in mind will help avoid pitfalls when dealing with similar improper integrals that mix polynomial growth and exponential decay.

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