integrals

Nice problems!

(1) can be written as $\int_0^\infty \frac{\sin^2 (ax)}{x^2}dx-\int_0^\infty \frac{\sin^2 (ax)}{1+x^2}$

The first integral is known and its value is $\frac{\pi}{2}a$ . The second can be calculated with contour integration, yielding $\frac{\pi}{4}\left[1-e^{-2a}\right]$ .

(3) note that $\frac{1}{1+\cosh x} = \frac{2}{e^x+2+e^{-x}}=\frac{2}{(e^{\frac{1}{2}x}+e^{-\frac{1}{2}x})^2}=\frac{2e^{-x}}{(1+e^{-x})^2}$

and since $\frac{t}{(1-t)^2}=\sum_{k=1}^\infty kt^k$ for $|t|<1$ , we can put $t=-e^{-x}$ to obtain:

$\frac{1}{1+\cosh x} = 2 \sum_{k=1}^\infty (-1)^{k+1} ke^{-kx}$

Also note that $\int_0^\infty x^n e^{-kx}dx = \frac{\Gamma(n+1)}{k^{n+1}}$ and $\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^n} = \left(1-2^{1-n}\right)\zeta(n)$ .

From all these we obtain:

$\int_0^\infty \frac{x^n}{1+\cosh x} dx = 2 \int_0^\infty \left[\sum_{k=1}^\infty (-1)^{k+1}ke^{-kx}\right]dx$

$= 2 \sum_{k=1}^\infty (-1)^{k+1}k \int_0^\infty x^ne^{-kx}\right]dx = 2 \sum_{k=1}^\infty (-1)^{k+1} \frac{\Gamma(n+1)}{k^n}$

$=2\left(1-2^{1-n}\right)\zeta(n)\Gamma(n+1)$

and when $n>1$ is a natural number, this is $2\left(1-2^{1-n}\right)n!\zeta(n)$

(2) Let $J(a,b)=\int_0^\infty \frac{e^{-bx^2}}{x^2+a^2} dx$ , with $a,b>0$ .

Then $\int_0^\infty \frac{e^{-bx^2}}{(x^2+a^2)^2}dx = -\frac{1}{2a}\frac{\partial J}{\partial a}$ , and we're looking for this integral, evaluated when $a^2=\frac{1}{2}$ , $b=1$ .

To calculate $J$ , note that:

$a^2 J - \frac{\partial J}{\partial b} = \int_0^\infty \frac{a^2 e^{-bx^2}}{x^2+a^2}dx+\int_0^\infty \frac{x^2 e^{-bx^2}}{x^2+a^2}dx = \int_0^\infty e^{-bx^2}dx = \frac{1}{2}\sqrt\frac{\pi}{b}$

So that:

$\frac{\partial}{\partial b}\left(e^{-a^2b} J\right) = -\frac{1}{2}\sqrt\frac{\pi}{b} e^{-a^2b}$

Now let $f(t) = \int_0^t e^{-x^2}dx$ .

It is easily checked that the indefinite integral with respect to $b$ of the right hand side of the previous equation is $-\frac{\sqrt\pi}{a} f(a\sqrt{b})+C$ , so:

$e^{-a^2b} J = -\frac{\sqrt\pi}{a} f(a\sqrt{b})+C(a)$

From the definition of $J$ it is obvious that the left hand side vanishes when $b=\infty$ , so the right hand side also vanishes. Thus, using $f(\infty)=\frac{\sqrt\pi}{2}$ , we get $C=\frac{\pi}{2a}$ . Therefore:

$J=\frac{\sqrt\pi e^{a^2 b}}{a}\left[\frac{\sqrt\pi}{2}-f(a\sqrt{b})\right]$

Now we differentiate with respect to $a$ , remembering that $\frac{d}{dt} f(t) = e^{-t^2}$ :

$\frac{\partial J}{\partial a} = \sqrt{\pi}e^{a^2 b}\frac{2a^2b-1}{a^2} \left[\frac{\sqrt\pi}{2}-f(a\sqrt{b})\right]-\frac{\sqrt{\pi b}}{a}$ .

Now, as we noted, $\int_0^\infty \frac{e^{-bx^2}}{(x^2+a^2)^2}dx = -\frac{1}{2a}\frac{\partial J}{\partial a}$ , so finally we have:

$\int_0^\infty \frac{e^{-bx^2}}{(x^2+a^2)^2}dx = \frac{\sqrt{\pi b}}{2a^2}-\sqrt{\pi}e^{a^2 b}\frac{2a^2b-1}{2a^3} \left[\frac{\sqrt\pi}{2}-f(a\sqrt{b})\right]$ .

Fortunately, when $a^2=\frac{1}{2}$ and $b=1$ the second term vanishes, and thus

$\int_0^\infty \frac{e^{-x^2}}{(x^2+\frac{1}{2})^2}dx = \sqrt{\pi}$