# integrals

• Oct 8th 2011, 07:47 PM
Random Variable
integrals
Don't hide your solutions.

1) $\int_{0}^{\infty} \frac{\sin^{2} (ax)}{x^{2}(1+x^{2})} \ dx \ \ a>0$

2) $\int_{0}^{\infty} \frac{e^{-x^{2}}}{\big(x^{2}+ \frac{1}{2}\big)^{2}} \ dx$

3) $\int_{0}^{\infty} \frac{x^{n}}{1+\cosh x} \ dx \ \ n \in \mathbb{N}, \ n > 1$
• Oct 10th 2011, 04:06 PM
Unbeatable0
Re: 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

$\int_0^\infty \frac{e^{-x^2}}{(x^2+\frac{1}{2})^2}dx = \sqrt{\pi}$
• Oct 10th 2011, 07:12 PM
Random Variable
Re: integrals
Surprisingly, problem 2 can also be solved by integrating by parts.

$\int_{0}^{\infty} \frac{e^{-x^{2}}}{(x^2 + \frac{1}{2})^{2}}} \ dx = \int_{0}^{\infty} \frac{e^{-x^{2}}}{x} \frac{x}{(x^{2}+\frac{1}{2})^{2}}} \ dx$

$= -\frac{1}{2} \frac{e^{-x^{2}}}{x(x^{2}+\frac{1}{2})^{2}} \Big|^{\infty}_{0} - \int_{0}^{\infty} \frac{e^{-x^{2}}}{x^{2}} \ dx$

$= \Big(-\frac{1}{2} \frac{e^{-x^{2}}}{x(x^{2}+\frac{1}{2})^{2}} + \frac{e^{-x^{2}}}{x} \Big) \Big|_{0}^{\infty} + 2 \int_{0}^{\infty} e^{-x^{2}} \ dx$

$= \frac{xe^{-x^{2}}}{x^{2}+\frac{1}{2}} \Big|^{\infty}_{0} + 2 \int_{0}^{\infty} e^{-x^{2}} = \sqrt{\pi}$
• Dec 13th 2011, 02:47 PM
uniquesailor
Re: integrals