5

How to solve this integral?

$$\int_{-\infty}^{\infty} {\rm exp}\left(\frac{i b t - a}{t^2 + 1}\right) - {\rm exp}\left(\frac{i b}{t} - \frac{1}{t^2}\right) {\rm d}t\\ {\rm with}\, a>0,b\in \mathbb{R},i^2=-1$$

The integral converges only if both summands are considered together. The integral originates from the comments in this post.

granular_bastard
  • 563

1 Answers1

3

Since the image part of the integrand is odd, then the given integral is real, $$I(a,b)=2\int\limits_0^\infty \left(\exp\left(-\dfrac a{1+t^2 }\right) \cos\left(\dfrac{bt}{1+t^2}\right)-\exp\left(-\dfrac1{t^2}\right)\cos\left(\dfrac bt\right)\right)\,\text dt.$$ $$I(a,b)=2I_1(a,b)+2I_2(b),\tag1$$ where $$I_1(a,b)=\int\limits_0^\infty \left(\exp\left(-\dfrac a{1+t^2 }\right) \cos\left(\dfrac{bt}{1+t^2}\right)-1\right)\,\text dt,\tag2$$ $$I_2(b)=\int\limits_0^\infty \left(1-\exp\left(-\dfrac1{t^2}\right) \cos\left(\dfrac bt\right)\right)\,\text dt = \pi\,\dfrac{|b|}2\operatorname{erf}\left(\dfrac{|b|}2\right)+\sqrt\pi \exp\left(-\dfrac{b^2}4\right)\tag3$$ (see also WA integration).

Second integral

Edit

Since $$\int_0^\infty \dfrac{\text dt}{(1+t^2)^m} = \dfrac{\sqrt\pi}2 \dfrac{\Gamma(m-1/2)}{\Gamma(m)},\tag4$$ First integral, b=0

then $$I_1(a,0)=\int\limits_0^\infty \left(\exp\left(-\frac a{1+t^2}\right)-1\right)\,\text dt =\dfrac{\sqrt\pi}2\sum\limits_{m=1}^\infty \dfrac{(-a)^m \Gamma(m-\frac12)}{m!(m-1)!},$$ First integral, summation, b=0 $$I_1(a,0)=-\dfrac\pi2\,a \exp\left(-\dfrac a2\right)\left(\operatorname I_0\left(\dfrac a2 \right)+\operatorname I_1\left(\dfrac a2 \right)\right).\tag5$$

Besides, $$\begin{align} &\int\limits_0^{\large\frac\pi2}\exp\left(-\dfrac a2 \cos2x\right) \left(\dfrac b2 \sin^2 2x\right)^m \dfrac{\text dx}{\cos^2 x}\\[4pt] &=\dfrac{\sqrt\pi}4 \left(\frac{2b}a\right)^m\,\Gamma\left(m-\dfrac12\right) \left((a+4m)\operatorname I_m\left(\frac a2\right)+a\operatorname I_{m+1}\left(\frac a2\right)\right), \end{align}\tag6$$ and the substitution $\;t=\tan x\;$ allows to right $$I_1(a,b)-I_1(a,0) =\int\limits_0^\infty \exp\left(-\dfrac a{1+t^2}\right) \left(\cos\left(\dfrac{bt}{1+t^2}\right)-1\right)\,\text dt$$ $$=\int\limits_0^{\large\frac\pi2}\exp\left(-a\cos^2 x\right) \left(\cos\left(\dfrac b2\sin 2x\right)-1\right) \dfrac{\,\text dx}{\cos^2 x}$$ $$=\exp\left(-\dfrac a2\right) \int\limits_0^{\large\frac\pi2}\exp\left(-\dfrac a2\cos2x\right) \left(\cos\left(\dfrac b2\sin 2x\right)-1\right) \dfrac{\,\text dx}{\cos^2 x}$$ $$=\exp\left(-\dfrac a2\right) \sum\limits_{m=1}^\infty \dfrac1{(2m)!}\left(-1\right)^m \int\limits_0^{\large\frac\pi2}\exp\left(-\dfrac a2\cos2x\right) \left(\dfrac b2\sin^2 2x\right)^m \dfrac{\,\text dx}{\cos^2 x}$$ $$=\dfrac{\sqrt\pi}4 \exp\left(-\dfrac a2\right) \sum\limits_{m=1}^\infty \dfrac{\Gamma(m-\frac12)}{(2m)!} \left(-\frac{2b}a\right)^m \left((a+4m)\operatorname I_m\left(\frac a2\right)+a\operatorname I_{m+1}\left(\frac a2\right)\right).\tag7$$ Formulas $(1),(3),(5),(7)$ provide the result as 1d series of the special functions.

Old version

Using the integral representation of the beta-function in the form of $$\int\limits_0^{\large \frac\pi2}\dfrac{\sin^{2k}2x}{\cos^2x}\,\text dx =\dfrac{\sqrt\pi\, \Gamma(k-\frac12)}{\Gamma(k)}\tag4$$ and substitution $\;t=\tan x,\;$ one can get $$I_1(0,b)=\int\limits_0^\infty \left(\cos\left(\dfrac{bt}{1+t^2}\right)-1\right)\,\text dt =\int\limits_0^{\large\frac\pi2} \left(\cos\left(\dfrac b2 \sin 2x\right)-1\right) \dfrac{\text dx}{\cos^2 x}$$ $$=\sum\limits_{k=1}^{\infty}\dfrac{(-1)^k}{(2k)!}\left(\dfrac b2\right)^{2k} \int\limits_0^{\large\frac\pi2} \dfrac{\sin^{2k} 2x\,\text dx}{\cos^2 x} =\sum\limits_{k=1}^{\infty}\dfrac{(-1)^k}{(2k)!}\left(\dfrac b2\right)^{2k} \dfrac{\sqrt\pi\, \Gamma(k-\frac12)}{\Gamma(k)},$$ $$I_1(0,b)=-\pi\,\dfrac{b^2}8\, \operatorname{_1F_2}\left(\dfrac12;\dfrac32,2;-\dfrac{b^2}{16}\right).\tag5$$ First integral, a=0

Similarly, applying the integral representation of the beta-function in the form of $$\int\limits_0^{\large \frac\pi2} \sin^{2k}2x \cos^{2m-2}x\,\text dx =\dfrac{2^{2k-1} \Gamma(k+\frac12) \Gamma(k+m-\frac12)}{\Gamma(2k+m)}\tag6$$ and substitution $\;t=\tan x,\;$ one can get $$I_1(a,b)-I_1(0,b)=\int\limits_0^\infty \left(\exp\left(-\dfrac a{1+t^2 }\right)-1\right) \cos\left(\dfrac{bt}{1+t^2}\right)\,\text dt$$ $$=\int\limits_0^{\large\frac\pi2} \left(e^{-a\cos^2x}-1\right) \cos\left(\dfrac b2 \sin 2x\right)\,\dfrac{\text dx}{\cos^2 x}$$ $$=\sum\limits_{m=1}^{\infty}\dfrac{a^m}{m!} \sum\limits_{k=0}^{\infty}\dfrac{(-1)^{m+k}}{(2k)!} \left(\dfrac b2\right)^{2k} \int\limits_0^{\large\frac\pi2} \sin^{2k} 2x \cos^{2m-2}x\,\text dx$$ $$=\sum\limits_{m=1}^{\infty}\dfrac{a^m}{m!} \sum\limits_{k=0}^{\infty}\dfrac{(-1)^{m+k}}{(2k)!} \left(\dfrac b2\right)^{2k} \dfrac{2^{2k-1} \Gamma(k+\frac12) \Gamma(k+m-\frac12)}{\Gamma(2k+m)}$$ $$I_1(a,b)-I_1(0,b)=\dfrac{\sqrt\pi}2\sum\limits_{m=1}^{\infty}\dfrac{(-a)^m\Gamma(m-\frac12)}{(m-1)!m!} \operatorname{_1F_2}\left(m-\frac12;\frac{m+1}2,\frac m2; -\frac1{16}b^2\right).\tag7$$ First integral

Formulas $(1),(3),(5),(7)$ provide the result as 1d series of the special functions.

Yuri Negometyanov
  • 28,835
  • I'll note that there was a (rejected) suggested edit indicating that the line above equation 7 has a typo: it should read $(b/2)^2$ instead of $(b/2)$. I did not check this, but I'll note it. If this is incorrect, respond and I'll delete this comment. – davidlowryduda Sep 19 '22 at 14:18
  • I specify that the typo is in eq.7 of the upper version (i.e. not old version). Instead of $2b$ there should be $b^2$. This error follows from a typo in the line before. There it should be $(b/2)^2$ instead of $b/2$. The other equations in the upper version are correct. My suggested edit was presumably rejected because it cannot be checked so quickly. – granular_bastard Sep 19 '22 at 22:15