Does this integral have a closed-form solution: $\int_b^{\infty} \frac{\cos(ax)}{1+x^2} dx$?

Question

I'm very rusty with calculus, and I was hoping someone would be willing to help me with the following definite integral:

$$\int_b^{\infty} \frac{\cos(ax)}{1+x^2} dx$$

$$b>0$$

Thanks in advance.

Answer 1

For $b = 0$ or $b = -\infty$ you can use contour integration. For other $b$ (assuming $a \neq 0$), you will have to do it numerically.

(If there were a formula in terms of $b$, you would have an elementary formula for an antiderivative of the integrand (just differentiate your formula with respect to $b$ and you'll recover the negative of the integrand!), but there is no elementary formula for an antiderivative of this integrand, I believe).

Answer 2

This will be a non elementary solution which leaves the chance for expanding closed form solutions in terms of special functions like the trigonometric integral function. This uses the cosine series representation. Also see this fun graph. Integration proof and formulas for integration.

Sum representation:

$$\mathrm{\int \frac{cos(ax)}{x^2+1}dx=\int \frac{1}{x^2+1}\sum_{n=0}^\infty \frac{(-1)^n a^{2n}x^{2n}}{(2n)!} dx= \sum_{n=0}^\infty \frac{(-1)^n a^{2n}}{(2n)!}\int \frac{x^{2n}}{x^2+1}dx= \sum_{n=0}^\infty \frac{(-1)^n a^{2n}x^{2n+1} \,_2F_1\left(1,n+\frac12,n+\frac 32,-x^2\right)}{(2n+1)!}+C=C+ \sum_{n=0}^\infty \frac{(-1)^n a^{2n}x^{2n+1}\left(n+\frac12\right)Ф\left(-x^2,1,n+\frac12\right)}{(2n+1)!}}$$

The hypergeometric function can be decomposed into a Lerch Transcendent function.

Closed form of your integral using dlmf 6.7.3, complex cosine definition, and you could also use partial fraction decomposition to derive formula 6.7.3 with the use of the Exponential Integral function with b>0. The substitution is based on this result:

$$\mathrm{\int_b^\infty \frac{cos(ax)}{x^2+1}dx=\frac12 \int_b^\infty \frac{e^{aix}}{x^2+1}dx+ \frac12 \int_b^\infty \frac{e^{-aix}}{x^2+1}dx= \frac{\pi}{2e^a}+\frac{1}{2}\big(i\, cosh(a)\, [Ci(a ( i-b)) - Ci(a (b + i)] + sinh(a)[Si(a (b + i)) +\, Si(a(i - b)]\big)= \frac{\pi}{2e^a}+\frac14 i e^a [Ei(-a(i b + 1) - Ei(a(i b- 1))] - \frac14 i e^{-a}[ Ei(a(1 - i b)) + Ei(a(i b+ 1))]}$$

Please correct me and give me feedback!