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I wanna calculate: $$\int_0^\infty e^{-tx}\cos(x)\cdot A \, dx$$

where $A=\frac 2 {\sqrt\pi} \int_0^\infty e^{-xu^2}\,du$

What I've done :

Rewrite the integral to $$\int_0^\infty e^{-tx} \cos(x)\cdot\frac{2}{\sqrt\pi} \int_0^\infty e^{-xu^2} \, du \,dx$$

$$=\int_0^\infty \left(\int_0^\infty e^{-tx} \cos(x)\cdot\frac{2}{\sqrt\pi} e^{-xu^2} \, du \right)\, dx$$

Substitute $w=\sqrt xu\Rightarrow dt=\sqrt x \,du$

We get: $$=\int_0^\infty \left(\int_0^\infty e^{-tx} \cos(x)\cdot\frac 2 {\sqrt\pi} e^{w^2}\, dw \right)\, dx$$

Any hint ?

Michael Hardy
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pinkpanther5
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  • integration by parts twice or $cos(x)=Re(e^{ix})$. Also, since $A$ is constant, is more useful put it outside the integral – Veridian Dynamics Feb 01 '17 at 19:43
  • @Basti: except that $A$ is a function of $x$. –  Feb 01 '17 at 19:46
  • Using that $A = \frac{1}{\sqrt{x}}$ the integral can be written $2\int_0^\infty e^{-y^2 t}\cos(y^2) {\rm d}y$. Using $2\cos(y^2) = e^{iy^2} + e^{-iy^2}$ this can again be written as $\int_0^\infty [e^{-y^2(t+i)} + e^{-y^2(t-i)}] {\rm d}y$ which are Gaussian integrals. See e.g. http://math.stackexchange.com/questions/374433/calculating-int-infty-infty-e-ax2eibxdx – Winther Feb 01 '17 at 20:01
  • @Winther: Thanks ! – pinkpanther5 Feb 01 '17 at 20:12

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We have $$I=\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-tx}\cos\left(x\right)\int_{0}^{\infty}e^{-xu^{2}}dudx $$ now take $u\sqrt{x}=v,\, du=dv/\sqrt{x} $. We get $$I=\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}\frac{e^{-tx}\cos\left(x\right)}{\sqrt{x}}\int_{0}^{\infty}e^{-v^{2}}dvdx=\int_{0}^{\infty}\frac{e^{-tx}\cos\left(x\right)}{\sqrt{x}} $$ $$=\textrm{Re}\left(\int_{0}^{\infty}\frac{e^{-x\left(t-i\right)}}{\sqrt{x}}\right)=\textrm{Re}\left(\frac{1}{\sqrt{t-i}}\int_{0}^{\infty}\frac{e^{-y}}{\sqrt{y}}\right)=\textrm{Re}\left(\frac{\sqrt{\pi}}{\sqrt{t-i}}\right)$$ and now you can calculate the integral using the trigonometric representation of complex numbers.

Marco Cantarini
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  • Using trigonometric identities relating $\cos(2x)$ to $\tan(2x)$ and $\cos(2x)$ to $\cos(x)$ the last result can be simplified down to something like $\frac{\sqrt{\pi}}{\sqrt{1+t^2}}\sqrt{\frac{1 + \frac{t}{\sqrt{1+t^2}}}{2}}$ – Winther Feb 01 '17 at 20:24