Trouble with $I(\alpha) = \int_0^{\infty} \frac{\cos (\alpha x)}{x^2 + 1} dx$

Question

I'm ultimately trying to solve $$I(\alpha) = \int_0^{\infty} \dfrac{\cos (\alpha x)}{x^2 + 1} dx$$

by using differentiation under the integral. I realize that this is most easily done using residues but I'm intending this problem to introduce my advanced calculus 2/differential equations students to some interesting techniques before they take real analysis.

Differentiating under the integral a first time leads to

$$I'(\alpha) = \int_0^{\infty} \dfrac{-x \sin (\alpha x)}{x^2 + 1} dx = - \dfrac{\pi}{2} + \int_0^{\infty} \dfrac{\sin (\alpha x)}{x(x^2 + 1)}dx$$

by making use of the Dirichlet integral and again to

$$I''(\alpha) = \int_0^{\infty} \dfrac{\cos (\alpha x)}{x^2 + 1} = I(\alpha)$$

To solve this second-order ODE we'll need two initial conditions. The integral for $I'(\alpha)$ leads to the incorrect result $I'(0) = 0$ but the rewritten version leads to the correct result of $I'(0) = -\dfrac{\pi}{2}$. I'm having trouble justifying this.

Any help or guidance is appreciated. I'll also settle for simpler arguments as to why $I'(0) \neq 0$.

Since $\int \frac{x}{x^2+1} = \log\sqrt{x^2+1}$ what you really have in the first integral is $0\cdot \infty $ as $\alpha\to 0$. You could use the substitution $t=\alpha x$ and get the right limit, however. — Ninad Munshi, Jan 06 '21 at 20:45
It's better to start from $J(\alpha):=\int_0^\infty\frac{\sin(\alpha x)dx}{x(x^2+1)}$. — J.G., Jan 06 '21 at 21:38
In case it is interesting from somebody, you can find several other questions about the same integral using SearchOnMath and Approach Zero. See also: How to search on this site? — Martin Sleziak, Jan 07 '21 at 09:23
Since this is about differentiation under integral sign, I'll mention examples which do not use complex analysis: How do you Evaluate $\int_{-\infty}^\infty \frac{\cos(x)}{x^2+1}dx$ Without Using Residue Calculus, Evaluate the integral $\int\limits_{-\infty}^\infty \frac{\cos(x)}{x^2+1}dx$. (Probably you can find also other posts about this.) — Martin Sleziak, Jan 07 '21 at 09:25
When you created this post, you tagged it "unif". This created a new tag on Math SE---usually we ask that users discuss the introduction of new tags before creating them. However, I presume that this was a typological error, and have removed the tag---I don't know which tag you meant to apply, so please add an appropriate tag if you deem it necessary. — Xander Henderson, Jan 07 '21 at 12:39

Chris · Accepted Answer · 2021-01-06T21:37:41.667

3

You are assuming that $$ \int_{0}^{\infty}\frac{\sin(\alpha x)}{x}dx= \frac{\pi}{2} $$ but if $\alpha=0$, then $$ \int_{0}^{\infty}\frac{\sin(\alpha x)}{x}dx=0 $$ So, the equality $$ \int_{0}^{\infty}\frac{−x\sin(αx)}{x^2+1}dx=-\frac{π}{2}+\int_{0}^{\infty}\frac{\sin(αx)}{x(x^2+1)}dx $$ is true iff $\alpha>0$.

edited Jan 06 '21 at 21:37

answered Jan 06 '21 at 21:10

Chris

140

Ahh, good point! Thanks! – BobaFret Jan 06 '21 at 21:16
2

In fact, iff $\alpha>0$, since if $\alpha<0$ then $\int_0^\infty\frac{\sin(\alpha x)}{x}dx=-\frac{\pi}{2}$. – J.G. Jan 06 '21 at 21:33
Oh, that's true. Thanks. – Chris Jan 06 '21 at 21:36

Trouble with $I(\alpha) = \int_0^{\infty} \frac{\cos (\alpha x)}{x^2 + 1} dx$

1 Answers1