Generalization of $\int_0^{\infty} e^{-x^2} \cosh x d x$?

Question

Recently I had just met a solution on the integral using the power series of $\cosh x$. I wish to generalize it by the result the Gaussian Integral:

$$\int_{-\infty}^{\infty} e^{-a(x+b)^2} d x=\sqrt{\frac{\pi}{a}}$$

First of all, I, by definition, express the integrand in terms of $e^x$.

$$ \begin{aligned} \int_0^{\infty} e^{-a x^2} \cosh (b x) d x = & \frac{1}{4} \int_{-\infty}^{\infty}\left(e^{-a x^2+b x}+e^{-a x^2-b x}\right) d x \\ = & \frac{1}{4} \int_{-\infty}^{\infty}\left(e^{-a\left(x-\frac{b}{2 a}\right)^2+\frac{b^2}{4 a}}+e^{-a\left(x+\frac{b}{2 a}\right)^2+\frac{b^2}{4 a}}\right) d x \\ = & \left.\frac{e^{\frac{b^2}{4 a}}}{4} \int_{-\infty}^{\infty} e^{-a\left(x-\frac{b}{2 a}\right)^2} d x+\int_{-\infty}^{\infty} e^{-a\left(x+\frac{b}{2 a}\right)^2} d x\right] \\ = & \frac{e^{\frac{b^2}{4 a}}}{4}\left(\sqrt{\frac{\pi}{a}}+\sqrt{\frac{\pi}{a}}\right)\\=&\frac{e^{\frac{b^2}{4 a}}}{2} \sqrt{\frac{\pi}{a}} \end{aligned} $$ where $a>0$ and $b$ is real.

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Am I correct? Is there any other method?

Your comments and alternative methods are highly appreciated.

Answer 1

Yet another (still harder) solution. For fixed $a>0$ let $$F(b)=\int\limits_0^\infty e^{-ax^2}\cosh (bx)\,dx$$ Then $${d \over db}F(b)=\int\limits_{0}^\infty xe^{-ax^2}\sinh (bx)\,dx\\ \overset{\rm by\ parts}{=} -{1\over 2a}e^{-ax^2}\sinh(bx)\bigg\vert_0^\infty +{b\over 2a}\int_0^\infty e^{-ax^2}\cosh(bx)\,dx\\ ={b\over 2a}F(b)$$ Thus $\displaystyle F(b)=F(0)\exp\left ({b^2\over 4a}\right).$ Next $$F(0)=\int\limits_0^\infty e^{-ax^2}\,dx={1\over \sqrt{a}} \int\limits_0^\infty e^{-x^2}\,dx={\sqrt{\pi}\over 2\sqrt{a}}$$

Answer 2

A possibly harder solution:

$$ \begin{align} \int_{0}^{\infty}\exp\left(-ax^{2}\right)\cosh\left(bx\right)dx &= \int_{0}^{\infty}\exp\left(-ax^{2}\right)\sum_{k=0}^{\infty}\frac{\left(bx\right)^{2k}}{\left(2k\right)!}dx \\ &= \sum_{k=0}^{\infty}\frac{b^{2k}}{\left(2k\right)!}\int_{0}^{\infty}x^{2k}\exp\left(-ax^{2}\right)dx \\ &= \sum_{k=0}^{\infty}\frac{b^{2k}}{\left(2k\right)!}\frac{\left(k-\frac{1}{2}\right)!}{2a^{k+\frac{1}{2}}} \tag{1}\\ &= \frac{1}{2\sqrt{a}}\sum_{k=0}^{\infty}\left(\frac{b^{2}}{a}\right)^{k}\frac{1}{\left(2k\right)!}\Gamma\left(k+\frac{1}{2}\right) \\ &= \frac{1}{2\sqrt{a}}\sum_{k=0}^{\infty}\left(\frac{b^{2}}{a}\right)^{k}\frac{1}{\left(2k\right)!}\frac{\sqrt{\pi}\left(2k\right)!}{4^{k}k!} \tag{2} \\ &= \frac{\sqrt{\pi}}{2\sqrt{a}}\sum_{k=0}^{\infty}\left(\frac{b^{2}}{4a}\right)^{k}\cdot\frac{1}{k!} \\ &= \frac{\sqrt{\pi}}{2\sqrt{a}}\exp\left(\frac{b^{2}}{4a}\right) \end{align} $$

where $(1)$ is derived from this page and $(2)$ is derived from this Wiki.

Answer 3

We are given $$ \int_{-\infty}^{\infty} e^{-a(x+b)^2} d x=\sqrt{\frac{\pi}{a}},\quad a>0,b\in\mathbb R . \tag0$$ First, generalize this to $$ \int_{-\infty}^{\infty} e^{-a(x+b)^2} d x=\sqrt{\frac{\pi}{a}}, \quad a>0,b\in\mathbb C . \tag1$$ To prove it, note that the LHS is an analytic function of $b$ [by Morera], and constant on the real line, therefore constant on $\mathbb C$.

Next, if $\beta$ is real, we have \begin{align} \int_{-\infty}^{\infty}\exp\left(-ax^{2}\right)\cos\left(\beta x\right)dx &= \operatorname{Re} \int_{-\infty}^{\infty}\exp\left(-ax^{2}\right)\exp(i\beta x)\;dx \\ &= \operatorname{Re} \int_{-\infty}^\infty\exp\left(-a\left(x-\frac{i\beta}{2a}\right)^2\right)\exp\left(\frac{-\beta^2}{4a}\right)\;dx \\ &= \exp\left(\frac{-\beta^2}{4a}\right)\sqrt{\frac{\pi}{a}} . \end{align} Again, both sides are analytic in $\beta$, and we get $$ \int_{-\infty}^{\infty}\exp\left(-ax^{2}\right)\cos\left(\beta x\right)dx =\exp\left(\frac{-\beta^2}{4a}\right)\sqrt{\frac{\pi}{a}}, \quad a>0, \beta\in \mathbb C . \tag3$$ Therefore, for $a>0, b \in \mathbb C$, \begin{align} \int_{-\infty}^{\infty}\exp\left(-ax^{2}\right)\cosh\left(b x\right)dx & =\int_{-\infty}^{\infty}\exp\left(-ax^{2}\right)\cos\left(i b x\right)dx \\ &= \exp\left(\frac{-(ib)^2}{4a}\right)\sqrt{\frac{\pi}{a}} = \exp\left(\frac{b^2}{4a}\right)\sqrt{\frac{\pi}{a}} \end{align}