MHB Convergence of Integrals: Exploring Methods and Challenges

[mathmari]

Hey!

I want to check if the following integrals converge or not.

1. $\int_0^{\infty}e^{-x}\log (1+x)dx$
2. $\int_0^{\infty}\sqrt{x}\cos (x^2)dx$

Do we have to calculate these integrals or do we have to use for example Direct comparison test? (Wondering)

For the second one I tried for example $|\sqrt{x}\cos(x^2)|\leq \sqrt{x}$. But then the integral of $\sqrt{x}$ diverges, so this is not useful, right? (Wondering)

 

[Euge]

Hi mathmari,

One need not to compute the integrals to check for convergence. For the first integral, since $\log(1 + x) \le x$ for all $x \ge 0$, then the integrand is dominated by $xe^{-x}$, which has a convergent integral: $\int_0^\infty xe^{-x}\, dx = 1$. So by direct comparison with $xe^{-x}$, the first integral converges.

The second integral actually converges. Its value is $0.5\sin(\pi/8)\,\Gamma(3/4)$ by contour integration. Like I've said, you don't need to compute the integral. Since the integrand alternates sign, the direct comparison test will not be applicable. First, by the $u$-substitution $u = x^2$, we write $\int_0^\infty \sqrt{x} \cos(x^2)\, dx = \int_0^\infty u^{-1/2}\cos u\, du$. for each integer $n \ge 0$, $\int_{n\pi}^{(n+1)\pi} u^{-1/2}\cos u\, du = (-1)^n\int_0^{\pi} (v + n\pi)^{-1/2}\cos(v)\, dv$ using the sub $v = u - n\pi$ and the identity $\cos(v + n\pi) = (-1)^n\cos v$. So it suffices to show that the series

$$\sum_{n = 0}^\infty (-1)^n \int_0^\pi (v + n\pi)^{-1/2}\cos v\, dv$$

converges. Note the integrals in the above series are nonnegative and decreasing in $n$. So if you can prove

$$\lim_{n\to \infty} \int_0^\pi (v + n\pi)^{-1/2}\cos v\, dv = 0$$

then you can claim by the alternating series test that the series converges (and hence the original $\int_0^\infty u^{-1/2}\cos u\, du$ converges).

 

[mathmari]

One need not to compute the integrals to check for convergence. For the first integral, since $\log(1 + x) \le x$ for all $x \ge 0$, then the integrand is dominated by $xe^{-x}$, which has a convergent integral: $\int_0^\infty xe^{-x}\, dx = 1$. So by direct comparison with $xe^{-x}$, the first integral converges.

I understand! (Happy)

First, by the $u$-substitution $u = x^2$, we write $\int_0^\infty \sqrt{x} \cos(x^2)\, dx = \int_0^\infty u^{-1/2}\cos u\, du$. for each integer $n \ge 0$, $\int_{n\pi}^{(n+1)\pi} u^{-1/2}\cos u\, du = (-1)^n\int_0^{\pi} (v + n\pi)^{-1/2}\cos(v)\, dv$ using the sub $v = u - n\pi$ and the identity $\cos(v + n\pi) = (-1)^n\cos v$. So it suffices to show that the series

$$\sum_{n = 0}^\infty (-1)^n \int_0^\pi (v + n\pi)^{-1/2}\cos v\, dv$$

converges. Note the integrals in the above series are nonnegative and decreasing in $n$. So if you can prove

$$\lim_{n\to \infty} \int_0^\pi (v + n\pi)^{-1/2}\cos v\, dv = 0$$

then you can claim by the alternating series test that the series converges (and hence the original $\int_0^\infty u^{-1/2}\cos u\, du$ converges).

When we substitute $u=x^2$ we have that $du=2xdx \Rightarrow du=2\sqrt{x}\sqrt{x}dx$.
Since $u^{\frac{1}{4}}=\sqrt{x}$, we have that $\frac{1}{2u^{\frac{1}{4}}}du=\sqrt{x}dx$.
Therefore, $$\int_0^{\infty}\sqrt{x}\cos (x^2)dx=\int_0^{\infty}\frac{1}{2u^{\frac{1}{4}}}\cos (u)du=\frac{1}{2}\int_0^{\infty}u^{-\frac{1}{4}}\cos (u)du$$

How do we get $\int_0^\infty u^{-1/2}\cos u\, du$ ? What have I done wrong? (Wondering)

 

[mathmari]

Note the integrals in the above series are nonnegative and decreasing in $n$. So if you can prove

$$\lim_{n\to \infty} \int_0^\pi (v + n\pi)^{-1/2}\cos v\, dv = 0$$

How do we know that the integrals in the above series are nonnegative and decreasing in $n$ ? (Wondering)

Could you give me a hint how we could prove that $\lim_{n\to \infty} \int_0^\pi (v + n\pi)^{-1/2}\cos v\, dv = 0$ ? (Wondering)

 

[Euge]

When we substitute $u=x^2$ we have that $du=2xdx \Rightarrow du=2\sqrt{x}\sqrt{x}dx$.
Since $u^{\frac{1}{4}}=\sqrt{x}$, we have that $\frac{1}{2u^{\frac{1}{4}}}du=\sqrt{x}dx$.
Therefore, $$\int_0^{\infty}\sqrt{x}\cos (x^2)dx=\int_0^{\infty}\frac{1}{2u^{\frac{1}{4}}}\cos (u)du=\frac{1}{2}\int_0^{\infty}u^{-\frac{1}{4}}\cos (u)du$$

How do we get $\int_0^\infty u^{-1/2}\cos u\, du$ ? What have I done wrong? (Wondering)

Yes, you're correct. Sorry, it was late when I posted. So the integrals in the series I wrote should instead be $\int_0^\infty (v + n\pi)^{-1/4}\cos v\, dv$ (the factor of $1/2$ can be put outside the summation sign).

Now write

$$\int_0^\pi (v + n\pi)^{-1/4}\cos v\, dv = \int_0^{\pi/2} (v + n\pi)^{-1/4}\cos v\, dv + \int_{\pi/2}^\pi (v + n\pi)^{-1/4}\cos v\, dv$$

and note

$$\int_{\pi/2}^\pi (v + n\pi)^{-1/4}\cos v\, dv = -\int_0^{\pi/2} (v + \frac{\pi}{2} + n\pi)^{-1/4}\sin v\, dv$$

The series of interest is then $0.5 \sum\limits_{n = 0}^\infty (-1)^n a_n - 0.5\sum\limits_{n = 0}^\infty (-1)^n b_n$, where

$$a_n = \int_0^{\pi/2} (v + n\pi)^{-1/4}\cos v\, dv \quad \text{and}\quad b_n = \int_0^{\pi/2} (v + \frac{\pi}{2} + n\pi)^{-1/4} \sin v\, dv$$

Since the sine and cosine are nonnegative in the interval $[0,\pi/2]$, both $a_n$ and $b_n$ are nonnegative and monotone decreasing. So it suffices to prove $\lim\limits_{n\to \infty} a_n = 0 = \lim\limits_{n\to \infty} b_n$. For then, you can claim convergence of the series $\sum (-1)^n a_n$ and $\sum (-1)^n b_n$ by the alternating series test, and hence the convergence of the difference $0.5\sum (-1)^n a_n - 0.5\sum (-1)^n b_n$.

 

[mathmari]

Since the sine and cosine are nonnegative in the interval $[0,\pi/2]$, both $a_n$ and $b_n$ are nonnegative and monotone decreasing.

Do we have that $a_n$ and $b_n$ are monotone decreasing because $n$ is in the denominator? (Wondering)

 

[Euge]

Basically, yes.

 

[mathmari]

Ah ok! I understood it! (Happy)

Could we also show the convergence using the direct comparison for the integral $\int_0^\infty u^{-1/2}\cos u\, du$ ? But which inequality could we use here? (Wondering)

 

[Euge]

Direct comparison requires the integrand to not change sign, right? The cosine changes sign over $[0,\infty)$, so direct comparison will not do with that integral.

 

[mathmari]

Direct comparison requires the integrand to not change sign, right? The cosine changes sign over $[0,\infty)$, so direct comparison will not do with that integral.

Ah ok. So, the only way is the alternating series test, isn't it? (Wondering)

But at my other post http://mathhelpboards.com/analysis-50/convergence-integrals-20731.html#post94229 at the integral 2. I use the direct comparison test although we have the cosine. So, is this wrong? (Wondering)

 

[Euge]

I think you were missing my point. In your other post, the cosine is multiplied by a damping function, making the integral absolutely convergent. Since $\int_0^\infty u^{-1/2}\cos u\, du $ is not absolutely convergent, but conditionally convergent, you'll have problems finding an upper and lower bound, for which both integrals converge.

 

[mathmari]

I think you were missing my point. In your other post, the cosine is multiplied by a damping function, making the integral absolutely convergent. Since $\int_0^\infty u^{-1/2}\cos u\, du $ is not absolutely convergent, but conditionally convergent, you'll have problems finding an upper and lower bound, for which both integrals converge.

I see. Thank you very much! (Smile)