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I am interested in evaluating the integral $$ I = \int_{-\infty}^\infty\frac{\cos(kx)}{\sqrt{x^2+a^2}}\exp\left(-k\sqrt{x^2 + a^2}\right)dx $$

I have no clue where to start. I tried making $x=a\sinh(t)$ substitution, and then I got: $$ I = \int_{-\infty}^\infty\cos(ka\sinh(t))\exp(-ka\cosh(t))dt $$ $$ I = Re\int_{-\infty}^\infty\exp\left[ka(i\sinh(t) - \cosh(t))\right]dt $$

Any hints?

PS.: That integral came from trying to calculate the electric field at coordinates $(0, 0, a)$ due to an infinite plane at $z=0$ with surface charge density of $\sigma(x, y) = \sigma_0\cos(kx)\cos(ky)$. Not sure if this helps. $$ E = \frac{\sigma_0}{4\pi\epsilon_0}\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{\cos(kx)\cos(ky)}{x^2 + y^2 + a^2}dxdy = \frac{\sigma_0}{4\epsilon_0}\int_{-\infty}^\infty\frac{\cos(kx)}{\sqrt{x^2+a^2}}\exp\left(-k\sqrt{x^2 + a^2}\right)dx $$

Where I used $$\int_{-\infty}^\infty\frac{\cos(kt)}{t^2 + b^2}dt = \frac{\pi}{b}e^{-kb}$$

to solve for the integral in y.

Sine of the Time
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kid
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  • Have you tried evaluating the $x$-$y$-integral using polar coordinates? – Dermot Craddock Jun 12 '25 at 18:17
  • Have you tried rewriting $\cos(kx)\cos(ky) = 0.5(\cos(kx + ky) + \cos(kx - ky))$? You should then be able to use a rotated $x-y$ coordinate system and separate the integrals. – sudeep5221 Jun 12 '25 at 18:38
  • Along with the polar coordinate suggestion, I suspect you may have to use a standard trigonometric formula for $\cos u\cos v$ and then change variables by rotating $\pi/4$. Aha. @sudeep made the same suggestion just as I was typing. – Ted Shifrin Jun 12 '25 at 18:40
  • @DermotCraddock I think won't help because the numerator will be $cos(krcos\theta))$ for example, or something like that. – kid Jun 12 '25 at 18:48
  • @sudeep5221 and TedShifrin That's actually really good idea! I'll try that right now!!!! – kid Jun 12 '25 at 18:49
  • Your derivation of the integral is not correct? Shouldn't it be $$\frac{1}{4\pi\epsilon_0}\int_{\mathbb{R}^2}\frac{{\color{red}a}\sigma(x,y)}{(x^2+y^2+a^2)^{\color{red}{3/2}}}dA$$ – Quý Nhân Jun 12 '25 at 19:40
  • I think it would be better to put the double integral in the title. That is, IMHO, the meaningful question. I don't know that the specific integral needs to be in a title, but I agree that a meaningful title would be preferable. – Ted Shifrin Jun 12 '25 at 19:52
  • I think I solved it. Thanks for the help. Hopefully it is correct. – kid Jun 12 '25 at 19:58
  • @kid $\vec{dE}(x,y)=\frac{\sigma(x,y)dA}{4\pi\epsilon_0|(-x,-y,a)^T|^3}(-x,-y,a)^{T} $ , also in this case $\sigma(x,y)=\sigma(-x,-y)$ , thus $$\vec{E}=\int_{\mathbb{R}^2}\vec{dE}(x,y)=\int_{\mathbb{R}^2}\frac{1}{2}(\vec{dE}(x,y)+\vec{dE}(-x,-y))=\int_{\mathbb{R}^2}\frac{\sigma(x,y)dA}{4\pi\epsilon_0|(x,y,a)^T|^3}(0,0,a)^{T}$$ – Quý Nhân Jun 12 '25 at 20:14
  • @QuýNhân Sorry, I erased my comment because I realized what went wrong, and 10 seconds later you had yours explaining. Yeah, the integral is wrong. Oh well. Should have been 3/2, there should also be an $a$. I guess I'll have to redo this...... At least I had fun solving an integral hah – kid Jun 12 '25 at 20:20

1 Answers1

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$$ I = \int_{-\infty}^\infty\frac{\cos(kx)}{\sqrt{x^2+a^2}}\exp\left(-k\sqrt{x^2 + a^2}\right)dx $$

which comes from $$ I = \frac{1}{\pi} \int_{-\infty}^\infty\int_{-\infty}^\infty \frac{\cos(kx)\cos(ky)}{x^2 + y^2 + a^2}dxdy $$

Apply rotation by $\pi/4$ in xy coordinates, (thank you so much sudeep5221 and Ted Shifrin), then: $$ I = \frac{1}{2\pi} \int_{-\infty}^\infty\int_{-\infty}^\infty \frac{\cos(kx\sqrt{2}) + \cos(ky\sqrt{2})}{x^2 + y^2 + a^2}dxdy $$

Separate the integrals, and, they're the same by swapping $x,y$ coordinates: $$ I = \frac{1}{\pi} \int_{-\infty}^\infty\int_{-\infty}^\infty \frac{\cos(kx\sqrt{2})}{x^2 + y^2 + a^2}dxdy = \frac{1}{\pi} \int_{-\infty}^\infty\int_{-\infty}^\infty \frac{\cos(ky\sqrt{2})}{x^2 + y^2 + a^2}dxdy $$

Integrating the variable without cossine, and using this answer gives: $$ I = \int_{-\infty}^\infty\ \frac{\cos(kx\sqrt{2})}{\sqrt{x^2 + a^2}}dx = 2K_0(k|a|\sqrt{2}) $$

Alternatively, integrating the variable with cossine, substituting $x = a\sinh t$, and using this definition gives: $$ I = \int_{-\infty}^\infty \frac{\exp\left(-k\sqrt{2}\sqrt{x^2 + a^2}\right)}{\sqrt{x^2 + a^2}}dx = 2\int_{0}^\infty \exp\left(-ka\sqrt{2}\cosh t\right)dt = 2K_0(k|a|\sqrt{2}) $$

kid
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