I need to compute the integral:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_{0}^{\infty}dx\frac{cos(x)}{cosh(x)}[/tex]

so obviously i need to use the residue theorem here and evaluate the contour integral of the complex function (cos(z)/cosh(z)) the singularity point here is [tex]z=i\pi/2[/tex]

the intervals i decided to do the are:

(0,0)->(R,0)->(R,pi)->(0,pi)->(0,0)

and then ofcourse we shall let R approach infinity.

so by the residue theorem we get that:

[tex]\int_c\frac{cos(z)}{cosh(z)}=2i\pi Res(cos(z)/cosh(z),z=i\pi/2)[/tex]

on the other hand, after some algebraic manipulations we get that:

z=x+iy [tex]cos(z)/cos(z)=(cos(x)cos(y)cosh(x)cosh(y)+sin(y)sin(x)sinh(y)sinh(x))/(cos^2(y)cosh^2(x)+sin^2(y)sinh^2(x))([/tex]

now to calculate the integral we have that:

the above integral with z equals (before letting R approach infinity):

[tex]\int_{0}^{R}\frac{cos(x)}{cosh(x)}dx+i\int_{0}^{\pi}\frac{(cos(R)cos(y)cosh(y)cosh(R)+sin(y)sinh(y)sin(R)sinh(R))}{cos^2ycosh^2R+sinh^2ysinh^2R}-\int_{R}^{0}\frac{cosh(\pi)cos(x)}{cosh(x)}dx+i\int_{\pi}^{0}cos(y)cosh(y)dy[/tex]

now i think that one of the above summed integrals is when R approaches zero approaches zero as well, to prove this would be quite long cause i think i need there l'ohpital rule, the second imaginary part shouldn't be too much of a hastle, by parts, i just want to see if this approach is the correct one, or is there a simpler appraoch that will make my life easier?

thanks in advance.

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# Homework Help: Computation of an integral (Use of Resdiue theorem).

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