# Provide a different method of proving:$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}]^2}dx={1\over 2}$

Accidentally founded this particular integral producing a rational number

I can't be for sure it is correct, so can one provide a proof of it.

$$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}]^2}dx={1\over 2}\tag 1$$

I found related to $(1)$ is this

Let enforce a substitution of $u=x+e^{\pi}$ then $du=dx$

$$\int_{-\infty}^{\infty}{1\over [\pi{u}^2+\pi^{1/3}]^2}du={1\over 2}\tag2$$

To avoid confusing with too much $\pi$ symbol, we write a general

$$\int_{-\infty}^{\infty}{1\over [A{u}^2+B]^2}du={1\over 2}\tag3$$

We could apply partial decomposition

$${au+b\over Au^2+B}+{cu+d\over (Au^2+B)^2}=1\tag4$$ then find a,b,c and d.

I found a general integral of

$$\int{dx\over(x^2+a^2)^2}={x\over2a^2(x^2+a^2)}+{1\over 2a^3}\tan^{-1}{\left(x\over a\right)}\tag5$$ I am sure this is enough to prove $(1)$

Question: what are other methods can we apply to prove $(1)?$

Perhaps the most elegant approach here is to use contour integration in the complex plane. Rewrite $Au^2+B$ as a difference of squares and decompose into partial fractions:

$$\frac{1}{(Au^2+B)^2}=\frac{1}{(\sqrt{A}u+\sqrt{B}i)^2(\sqrt{A}u-\sqrt{B}i)^2}$$

$$=-\frac{1}{4B}\left(\frac{1}{(\sqrt A u + \sqrt{B}i)^2}+\frac{1}{(\sqrt A u - \sqrt{B}i)^2}-\frac{i/\sqrt{B}}{\sqrt A u + \sqrt{B}i}+\frac{i/\sqrt{B}}{\sqrt A u - \sqrt{B}i}\right)$$

Now we use the Residue theorem, which says, that an integral over a closed curve (we cap off the real line with an infinite half-circle over the upper half plane, where the integrand tends to zero anyway, so this addition doesn't change the value of the integral) equals the coefficients of poles of the $1/u$ shape, which are found inside the encircled area (upper plane in this case), multiplied by $2\pi i$. In our case, the only pole of the first order with positive imaginary part is the last term (with pole at $u_0=\sqrt{B/A}i$). The first two poles are quadratic and don't count and the third is on the lower half-plane and is not encircled by the contour. Rewrite the term just for clarity as

$$\color{red}{-\frac{1}{4B}\frac{i/\sqrt{B}}{\sqrt{A}}}\frac{1}{u-u_0}$$

and apply the residue theorem to the integral:

$$\int_{-\infty}^\infty\frac{1}{(Au^2+B)^2}du =$$ $$=\oint_{C}\frac{1}{(Au^2+B)^2}du=2\pi i \left(\frac{-1}{4B}\frac{i/\sqrt{B}}{\sqrt{A}}\right)=\frac{\pi}{2 \sqrt{AB^3}}=\frac{\pi}{2\sqrt{\pi^2}}=\frac12$$

• Lovely stuff, +1 – Kevin Jan 6 '17 at 14:16

Since holds $$I\left(a,b\right)=\int_{0}^{\infty}\sin\left(az\right)e^{-bz}dz=\frac{a}{a^{2}+b^{2}}$$ we have $$\frac{\partial}{\partial b}I\left(a,b\right)=-\int_{0}^{\infty}z\sin\left(az\right)e^{-bz}dz=-\frac{2ab}{\left(a^{2}+b^{2}\right)^{2}}$$ so \begin{align} I= & \int_{-\infty}^{\infty}\frac{du}{\left(Au^{2}+B\right)^{2}} \\ = & \frac{1}{2\sqrt{AB}}\int_{-\infty}^{\infty}\frac{1}{u}\int_{0}^{\infty}z\sin\left(\sqrt{A}uz\right)e^{-\sqrt{B}z}dzdu \\ = & \frac{1}{2\sqrt{AB}}\int_{0}^{\infty}ze^{-\sqrt{B}z}\int_{-\infty}^{\infty}\frac{1}{u}\sin\left(\sqrt{A}uz\right)dudz \\ = & \frac{1}{2\sqrt{AB}}\int_{0}^{\infty}ze^{-\sqrt{B}z}\int_{-\infty}^{\infty}\frac{1}{v}\sin\left(v\right)dvdz \end{align} and the last integral in the RHS is well known. So $$I=\frac{\pi}{2\sqrt{AB}}\int_{0}^{\infty}ze^{-\sqrt{B}z}dz=\frac{\pi}{2\sqrt{AB^{3}}}\int_{0}^{\infty}we^{-w}dw=\color{red}{\frac{\pi}{2\sqrt{AB^{3}}}}.$$

We can also apply reduction formula to $(3)$ as: $$\int \frac {1}{(au+b)^n} du = \frac {2n-3}{2b (n-1)}\int \frac{1}{(au+b)^{n-1}} du + \frac {u}{2b (n-1)(au^2+b)^{n-1}}$$ With $a=\pi , b=\sqrt [3]{\pi}, n=2$. The integral can then be solved by substituting $v=\sqrt [3]{\pi}u$.

Starting with, $$\int_{-\infty}^{\infty}{1\over [A{u}^2+B]^2}du$$ you could let $u=\sqrt{\frac{B}{A}} \tan(\theta)$ so $du=\sqrt{\frac{B}{A}} \sec^2(\theta)d \theta.$ Using $1+\tan^2(x)=\sec^2(x)$, this gives, $$\sqrt{\frac{B}{A}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2(\theta)}{(B\tan^2(\theta)+B)^2}d\theta=\sqrt{\frac{1}{AB^3}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta)d\theta.$$ Now using $cos^2(x)=\frac{1}{2}(cos(2x)+1)$, we have, $$\frac{1}{2\sqrt{AB^3}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(2\theta)+1 d\theta=\frac{1}{2\sqrt{AB^3}}(\frac{\sin(2\theta)}{2} +\theta)|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=\frac{\pi}{2\sqrt{AB^3}}.$$ Plugging in $A=\pi$ and $B=\pi^{\frac{1}{3}}$ gives the desired result.

• Maybe it is interesting to see that your calculations can be used to prove $(5)$. – Marco Cantarini Jan 6 '17 at 9:21

To avoid confusion, $a = A = \pi$ and $b = B = \pi ^{1/3}$

Let $z=\frac{a u^{2}}{b}$ \begin{align} \int\limits_{-\infty}^{\infty} \frac{1}{(a u^{2} + b)^{2}} du &= 2 \int\limits_{0}^{\infty} \frac{1}{(a u^{2} + b)^{2}} du \\ &= \frac{2}{b^{2}} \int\limits_{0}^{\infty} \frac{1}{(\frac{a}{b} u^{2} + 1)^{2}} du \\ &= \frac{1}{b^{2}} \sqrt{\frac{b}{a}} \int\limits_{0}^{\infty} \frac{z^{-1/2}}{(z+1)^{2}} dz \\ &= a^{-1/2} b^{-3/2} \mathrm{B}\left(\frac{1}{2},\frac{3}{2} \right) \\ &= \frac{1}{\pi} \mathrm{B}\left(\frac{1}{2},\frac{3}{2} \right) \\ &= \frac{1}{\pi} \frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)} \\ &= \frac{1}{2} \end{align}

Note that we used the following integral definition of the beta function $$\mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{z^{x-1}}{(z+1)^{x+y}} dz$$

Let us consider the Fourier Cosine Transform:

$$\mathcal{F}_{c} (f(x))= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) \cos(xy) \ dx ,$$ and the integral inner product on $L^2(\mathbb{R})$ which is $$\langle f(x),g(x) \rangle= \int_{-\infty}^{\infty} f(x)g(x) \ dx.$$

It turns out that $\mathcal{F}_{c}$ is a unitary operator, which means $$\langle f(x),g(x) \rangle=\langle \mathcal{F_c}(f(x)),\mathcal{F_c}(g(x)) \rangle$$ which can be seen by the Fourier Inversion theorem, or with this (heuristic) proof below (similar to Wikipedia's proof).

$$\langle \mathcal{F_c}(f(x)),\mathcal{F_c}(g(x)) \rangle = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)g(z) \cos(zy) \cos(xy) \ dz \ dx \ dy.$$ Reversing the order of integration, $$\frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)g(z) \cos(zy) \cos(xy) \ dy \ dz \ dx = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)g(z) \delta(x-z) \ dz \ dx$$ $$= \int_{-\infty}^{\infty} f(x)g(x) \ dx.$$

Now let us consider $$f(x)=g(x)= e^{-\pi^{\frac{1}{6}} |x|}$$ It can be shown by integration by parts that $$\mathcal{F}_c(f(x)) = \frac{1}{\sqrt{2 \pi}}\frac{2 \pi^{\frac{1}{6}}}{\pi^{\frac{1}{3}} +y^2}.$$

$$\langle f(x),g(x) \rangle= \int_{-\infty}^{\infty} e^{-2\pi^{\frac{1}{6}} |x|} \ dx=2\int_{0}^{\infty} e^{-2\pi^{\frac{1}{6}} x} \ dx= \frac{1}{\pi^{\frac{1}{6}}}.$$

On the other hand, exploiting the unitary property, this is equal to $$\langle \mathcal{F_c}(f(x)),\mathcal{F_c}(g(x)) \rangle = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{4\pi^{\frac{1}{3}}}{(\pi^{\frac{1}{3}} +y^2)^2} \ dy.$$ This implies by rearranging constants $$\frac{1}{2} \sqrt{\pi} = \int_{-\infty}^{\infty} \frac{1}{(\pi^{\frac{1}{3}} +y^2)^2} \ dy.$$

Finally, let $y=\sqrt{\pi} (u+e^{\pi}),$ so that $dy = \sqrt{\pi} du.$ The final result, upon rearranging constants is

$$\frac{1}{2}= \int_{-\infty}^{\infty} \frac{1}{\left(\pi(u+e^\pi)^2 + \pi^{\frac{1}{3}}\right)^2} \ du.$$