# Integrate $\int_{-\infty}^{\infty}\exp\left(-\frac{\pi^2t(2x+1)^2}{2c^2}\right)\cos\left(\frac{(2x+1)\pi y}{c}\right)\exp(-2\pi i kx)dx$

By the poisson summation formula we have:

$$\frac{1}{c}\sum\limits_{k=-\infty}^{\infty} \exp\left(-\frac{\pi^2t(2k+1)^2}{2c^2}\right)\cos\left(\frac{(2k+1)\pi y}{c}\right)\\=\frac{\exp(\frac{-y^2}{2t})}{\sqrt{2\pi t}}\left\lbrace 1+2\sum_{k=1}^{\infty}(-1)^k\exp(-\frac{-k^2c^2}{2t})\cosh(\frac{k y c}{t})\right\rbrace,$$

where $c,t>0$ is a constant and $y$ is also a fixed real number.

The Poisson summation formula says $$\sum_{k=-\infty}^{\infty}f(k)=\sum_{k=-\infty}^{\infty}\hat{f}(k),$$

where $\hat{f}(w):=\int_{-\infty}^{\infty}f(t)\exp(-2\pi i wt)dt$ is the Fourier transform.

I have problems in calculating the fourier transfrom of this particular function, i.e. to calculate the following integral:

$$\int_{-\infty}^{\infty}\exp\left(-\frac{\pi^2t(2x+1)^2}{2c^2}\right)\cos\left(\frac{(2x+1)\pi y}{c}\right)\exp(-2\pi i kx)dx,$$

where $k\in \mathbb{Z}$.

I tried to use $\cos(x)=\frac{1}{2}\cdot(e^{ix}+e^{-ix})$, but then I still couldn't calculate the value of the integral. I would really appreciate if someone can tell me the idea how to solve it.

Regards

$$I:=\int_{-\infty}^{\infty}\exp(-\frac{\pi^2t(2x+1)^2}{2c^2})\cos(\frac{(2x+1)\pi y}{c})\exp(-2\pi i kx)dx.$$ Let $\alpha \equiv \pi^2 c^{-2}t/2$ and a change of variables $2x+1=u$. Also, note $\Re (e^{ix}) = \cos x$. We then obtain $$\Re\bigg[ \frac{1}{2} \int_{-\infty}^\infty e^{-\alpha u^2} e^{{i u \pi y c^{-1}}} e^{-\pi i k(u-1)}du\bigg].$$ We now define $b \equiv \pi y c^{-1}$ and $\omega\equiv-\pi k$ to obtain $$\Re\bigg[\frac{1}{2}e^{-i\omega}\int_{-\infty}^\infty e^{-\alpha u^2}e^{iub}e^{i\omega u}du\bigg]=\Re\bigg[\frac{1}{2}e^{-i\omega}\int_{-\infty}^\infty e^{-\alpha u^2}e^{iu(b+\omega)}du\bigg].$$ Let $\zeta\equiv b+ \omega$ to obtain $$\Re\bigg[\frac{1}{2}e^{-i\omega}\int_{-\infty}^\infty e^{-\alpha u^2}e^{i u \zeta}du\bigg]$$

Recall $\omega=-\pi k, b=\pi y c^{-1},\alpha \equiv \pi^2 c^{-2}t/2$. Note, to get this result We need to Fourier Transform the Gaussian which requires complex variables. Let's re-define $x\equiv u$ and $\lambda\equiv \alpha$. Here it is

${\bf{Proof \ of \ Fourier \ Transform \ of \ Gaussian \ in \ a \ general \ form \ is \ here:}}$, Computing the Fourier transform of the Gaussian $$\int_{-\infty}^{\infty} e^{i\zeta x-\lambda x^2}dx.$$ We do this by writing $$\int_{-\infty}^\infty e^{i\zeta x} e^{-\lambda x^2}dx=\int_{-\infty}^{\infty} \cos(\zeta x) e^{-\lambda x^2}dx +i\int_{-\infty}^{\infty}\sin(\zeta x) e^{-\lambda x^2}dx=\int_{-\infty}^{\infty} \cos(\zeta x) e^{-\lambda x^2}dx$$ where I used $$\int_{-\infty}^{\infty}e^{-\lambda x^2} \sin(\zeta x) dx=0$$ since this integrand is odd, thus integration over symmetric bounds is zero. The integral is even so we can write $$\int_{-\infty}^{\infty} \cos(\zeta x) e^{-\lambda x^2}dx.=2\int_{0}^{\infty} e^{-\lambda x^2}\cos( \zeta x).$$ I now integrate this function using contour integration and using a Rectangular contour, standard for Gaussian complex integrals, see Churchill- Complex Variables. A diagram is shown below

We use the complex function $f(z)=e^{-\lambda z^2}$ which is an entire function thus by the Cauchy-Goursat theorem we have $$0=\oint_C f(z)dz=\int_{C_1}+\int_{C_2} +\int_{C_3} + \int_{C_4}.$$ The integral of $f(z)$ along $C_1$ due to the parametrization $z=x, dz=dx$, we obtain $$\int_{C_1} e^{-\lambda z^2} dz= \int_{-R}^R e^{-\lambda x^2}dx = 2 \int_0^R e^{-\lambda x^2}dx.$$ The integral of $f(z)$ along $C_3$ is parametrized by writing $z=x+i\zeta/(2\lambda)$, $z^2=x^2-\zeta^2/(4\lambda) +i\zeta x/\lambda$ $$\int_{C_3} e^{-\lambda z^2}dz= -\int_{-R}^{R }e^{-\lambda(x^2-\frac{\zeta^2}{4\lambda^2}+\frac{i\zeta x}{\lambda})} dx=-e^{\frac{\zeta^2}{4\lambda^2}} \int_{-R}^{R} e^{-\lambda x^2} e^{-i\zeta x}dx.$$ We simplify this by using $e^{-i\zeta x}=\cos \zeta x - i\sin \zeta x$, thus we obtain $$\int_{C_3}=-e^{\frac{\zeta^2}{4\lambda^2}}\bigg( \int_{-R}^{R} e^{-\lambda x^2}\cos (\zeta x) dx -i\int_{-R}^{R} \sin (\zeta x) e^{-\lambda x^2} dx \bigg)=-e^{\frac{\zeta^2}{4\lambda^2}} \int_{-R}^{R} e^{-\lambda x^2}\cos (\zeta x) dx=-2e^{\frac{\zeta^2}{4\lambda^2}}\int_{0}^{R} e^{-\lambda x^2}\cos (\zeta x) dx.$$ Note, again the integral of $e^{-\lambda x^2} \sin(\zeta x)$ is zero since it is odd function integrated over symmetric bounds. Thus we conclude that the integral along contour $C_3$ is given by $$\int_{C_3}f(z)dz=-2e^{\frac{\zeta^2}{4\lambda^2}}\int_{0}^{R} e^{-\lambda x^2}\cos (\zeta x) dx.$$ Now integrating along $C_2$ and $C_4$ by parametrizing using $z=\pm R +iy$, $z^2=R^2-y^2\pm 2iR\lambda y$, $dz=idy$ and noting that $-\frac{\zeta}{2\lambda} \leq y \leq \frac{\zeta}{2\lambda}$, we obtain $$\int_{C_2} f(z) dz=\int_{0}^{\frac{\zeta}{2\lambda}} e^{-\lambda R^2} e^{\lambda y^2} e^{-2iR\lambda y}idy.$$ The integral along $C_4$ yields $$\int_{C_4} f(z)dz= -\int_{0}^{\frac{\zeta}{2\lambda}} e^{-\lambda R^2} e^{\lambda y^2} e^{2iR\lambda y } idy.$$ Combining these we obtain $$\int_{C_2} +\int_{C_4}= ie^{-\lambda R^2}\int_{0}^{\frac{\zeta}{2\lambda}} e^{\lambda y^2} \big(e^{-2iR\lambda y}-e^{2iR\lambda y} \big)dy=2e^{-\lambda R^2} \int_{0}^{\frac{\zeta}{2\lambda}} e^{\lambda y^2} \sin(2R\lambda y) dy$$ where I used a sine identity. We are now ready to calculate the integral since we have calculated the contributions along all 4 contour paths. Thus using these results and Cauchy-Goursat theorem we obtain $$0=\oint_C f(z)dz=2\int_{0}^{R} e^{-\lambda x^2}dx-2e^{\frac{\zeta^2}{4\lambda}} \int_{0}^{R} e^{-\lambda x^2} \cos (\zeta x) dx + 2e^{-\lambda R^2} \int_{0}^{\frac{\zeta}{2\lambda}} e^{\lambda y^2}\sin(2R\lambda y)dy=0.$$ Simplfying this and arranging for the integral we want to calculate we obtain $$\int_{0}^{R} e^{-\lambda x^2} \cos (\zeta x) dx= e^{-\frac{\zeta^2}{4\lambda}} \int_{0}^{R} e^{-\lambda x^2} dx +e^{-(\lambda R^2+\frac{\zeta^2}{4\lambda})}\int_{0}^{\frac{\zeta}{2\lambda}} e^{\lambda y^2} \sin(2R\lambda y )dy.$$ We now prove that the integral $$e^{-(\lambda R^2+\frac{\zeta^2}{4\lambda})}\int_{0}^{\frac{\zeta}{2\lambda}} e^{\lambda y^2} \sin(2R\lambda y )dy \to 0 \ \text{as} \ R\to \infty.$$ We do this by using $$\bigg|\int_{0}^{\frac{\zeta}{2\lambda}} e^{\lambda y^2}\sin(2R\lambda y)dy\bigg| \leq \int_{0}^{\frac{\zeta}{2\lambda}} e^{\lambda y^2} dy$$ which is just a fixed number since $\zeta/(2\lambda)$ is fixed. Thus we obtain $$\bigg| e^{-(\lambda R^2+\frac{\zeta^2}{4\lambda})}\int_{0}^{\frac{\zeta}{2\lambda}} e^{\lambda y^2} \sin(2R\lambda y )dy \bigg| \leq e^{-(\lambda R^2+\frac{\zeta^2}{4\lambda})} \int_{0}^{\frac{\zeta}{2\lambda}} e^{\lambda y^2} dy \ \to 0 \ \text{as} \ R\to\infty \ \text{since} \ e^{-\lambda R^2} \to 0.$$ Thus we can simplify and obtain $$\int_{0}^{R} e^{-\lambda x^2} \cos (\zeta x) dx= e^{-\frac{\zeta^2}{4\lambda}} \int_{0}^{R} e^{-\lambda x^2} dx.$$ Taking the limit as $R \to \infty$ on both sides and extending bounds of integration to $-R$ to $R$ since the functions are even we obtain $$\lim_{R\to\infty} \frac{1}{2} \int_{-R}^{R} e^{-\lambda x^2} \cos (\zeta x) dx=\lim_{R\to\infty} \frac{1}{2} e^{-\frac{\zeta^2}{4\lambda}} \int_{-R}^{R} e^{-\lambda x^2} dx.$$ Thus we can see that $$\int_{-\infty}^{\infty} e^{-\lambda x^2} \cos ( \zeta x) dx= e^{-\frac{\zeta^2}{4\lambda}} \int_{-\infty}^{\infty} e^{-\lambda x^2} dx.$$ We have reduced the integral we wanted to calculate to a real Gaussian integral which is given by $$\int_{-\infty}^{\infty} e^{-\lambda x^2} dx=\sqrt{\frac{\pi}{\lambda}}.$$ ${\bf{This \ concludes \ THE\ FINAL\ RESULT\ by\ showing \ Fourier\ transform \ of \ the \ Gaussian}}$ is given by $$\int_{-\infty}^{\infty} e^{i\zeta x-\lambda x^2} dx= \int_{-\infty}^{\infty} \cos(\zeta x) e^{-\lambda x^2} dx=\sqrt{\frac{\pi}{\lambda}} e^{-\frac{\zeta^2}{4\lambda}}, \ \lambda > 0.$$ and your result is $$\Re\bigg[\frac{1}{2}e^{-i\omega}\sqrt{\frac{\pi}{\lambda}} e^{-\frac{\zeta^2}{4\lambda}}\bigg]$$

Re call $\zeta=\omega+b, \ \omega=-\pi k, b=\pi y c^{-1},\lambda \equiv \pi^2 c^{-2}t/2$. Use $e^{-i\omega}=\cos \omega-i\sin \omega$ and just take the real part! I think you can take it from here. Let me know if i made errors, its 3AM and I am tired. but this is the general idea. Fourier transform of gaussian

• A well deserved +1. – JamalS Apr 29 '14 at 9:22
• @Braten thank you for the interesting problem. I fixed some small typos. Glad it helped. – Jeff Faraci Apr 29 '14 at 16:22
• @user1997744 Thank you my friend.. – Jeff Faraci Apr 29 '14 at 16:28
• So elegant . +1 – user85798 Apr 29 '14 at 20:47
• Wow! Cool Jeff! I online on Math SE only to vote up your answers. :D – Tunk-Fey May 14 '14 at 4:59