Integrating using Laplace Transforms

$$\int_{0}^\infty {\cos(xt)\over 1+t^2}dt$$

I'm supposed to solve this using Laplace Transformations.

I've been trying this since this morning but I haven't figured it out. Any pointers to push me in the right direction?

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What do you mean by "solve"? You mean "calculate"? – Martin Argerami Nov 24 '12 at 16:41

Let $f(x)$ denote the integral, and assume temporarily that $x > 0$. This makes no difference since $f(x)$ is even by definition. Then its Laplace transform $\mathcal{L}f(s)$ defines a continuous function on $s \in (0, \infty)$ (in fact, it defines an analytic function on $\Re(s) > 0$). Thus we may assume further that $s \neq 1$ and then rely on the continuity argument. Then

\begin{align*} \mathcal{L}f(s) &= \int_{0}^{\infty} f(x)e^{-sx} \, dx = \int_{0}^{\infty} \int_{0}^{\infty} \frac{\cos(xt)}{1+t^2} e^{-sx} \, dtdx \\ &\stackrel{*}{=} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\cos(xt)}{1+t^2} e^{-sx} \, dxdt \\ &= \int_{0}^{\infty} \frac{1}{1+t^2} \frac{s}{s^2+t^2} \, dt \\ &= \frac{s}{1-s^2}\int_{0}^{\infty} \left( \frac{1}{s^2+t^2} - \frac{1}{1+t^2} \right) \, dt \\ &= \frac{s}{1-s^2} \frac{\pi}{2} \left( \frac{1}{s} - 1 \right) = \frac{\pi}{2} \frac{1}{1+s}. \end{align*}

Here, the change of order of integration $(*)$ is justified by the dominated convergence theorem. Though we proved this for $s \neq 1$, it remains valid by continuity argument as mentioned above. Then by the uniqueness of the Laplace transform, we find that

$$f(x) = \frac{\pi}{2}e^{-x}.$$

Therefore we have

$$\int_{0}^{\infty} \frac{\cos(xt)}{1+t^2} \, dx = \frac{\pi}{2} e^{-\left|x\right|}.$$

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Your answer is derived only for $x \geq 0$, but the integral makes sense for all $x$, so a more complete answer is $(\pi/2)e^{-|x|}$. – KCd Nov 24 '12 at 17:02
@KCd, Thanks. Of course I knew that, but I forgot mentioning that. – Sangchul Lee Nov 24 '12 at 17:06

Laplace approach : $$I= \int\limits_{0}^\infty {\cos(xt)\over 1+t^2}\,\mathrm dt=\left[ \int\limits_{0}^\infty {\cos(xt)\over 1+t^2} e^{-st}\,\mathrm dt \right]_{s=0} =\mathcal{L}\left[ {\cos(xt)\over 1+t^2} \right]_{s=0}$$ $$f(t)=\frac{\cos(xt)}{1+t^2}$$ $$(1+t^2)f(t)=\cos(xt)$$ $$\downarrow\mathcal{L}$$ $$\frac{\mathrm d^2}{\mathrm ds^2}F(s)+F(s)=\frac{s}{s^2+x^2}$$ We have second order differential equation, after we find $F(s)$ we put zero instead of $s$.

Where we have used : $$\boxed{\mathcal{L} \big\{ t^nf(t)\big\}=(-1)^n \dfrac{\mathrm d^n}{\mathrm ds^n}F(s)}$$ $$\boxed{\mathcal{L}\big\{\cos(wt) \big\} = \dfrac{s}{s^2+w^2}}$$

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I don't understand your edit @Brian M. Scott. – Elements in Space Jan 4 '13 at 18:25
@UnkleRhaukus: The OP’s notation for that step was perfectly acceptable. I consider it inappropriate to make unnecessary changes in someone else’s post, so I restored the original notation. (I was making that change to your edit and restoring the OP’s $d$’s when I discovered that someone had approved the edit while I was working on it.) – Brian M. Scott Jan 4 '13 at 18:32
I made those edits because it thought they made the post clearer more precise, and easier to read. Do you think restoring that "notation" is necessary or beneficial ? – Elements in Space Jan 4 '13 at 18:49

We can use residue theorm to solve this problem

$$\int_{-\infty}^{+\infty} cos(ax) f(x)dx=Real \left[ \oint_{c^+} e^{iaz} f(z)dz \right] = Real \left[ 2\pi i \sum_{i=1}^n R_i^+\right]$$ $$c^+ : Upper \space half \space plane \rightarrow uhp$$ $$c^- : Lower \space half \space plane \rightarrow lhp$$ $$R_i^+=Residue(e^{iaz}f(z),z=z_i) \space; \space z_i \space are \space singularities \space in \space c^+$$ $$I= \int_{0}^\infty {\cos(xt)\over 1+t^2}dt=\frac{1}{2}\int_{-\infty}^\infty {\cos(xt)\over 1+t^2}dt$$ $$I=\frac{1}{2}Real \left[ \oint_{c^+} \frac{e^{ixz}}{1+z^2} dz \right] =\frac{1}{2} Real \left[ 2\pi i .R_1 \right]$$ $$Singularities : 1+z^2=0 \rightarrow \begin{cases} z_1=i &\mbox{Acceptable (uhp)} \\ z_2=-i & \mbox{Ineligible (lhp) } \end{cases}$$

$$R_1 = Residue(\frac{e^{ixz}}{1+z^2},z=i)=\lim_{z \rightarrow i} \frac{(z-i)e^{ixz}}{(z-i)(z+i)}=\frac{e^{-x}}{2i}$$ so: $$I=\frac{1}{2} Real \left[ 2\pi i .R_1 \right]=\frac{1}{2} Real \left[ 2\pi i .\frac{e^{-x}}{2i} \right] =\frac{\pi}{2} e^{-x}$$

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