Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form? ∫0∞cos⁡x1+x2dx

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Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form?  ∫0∞cos⁡x1+x2dx

Bob Huerta · Accepted Answer

Let F(y)=∫0∞sin⁡xyx(1+x2)dx for y&amp;gt;0  Show that  F(y)−F(y)+π2=0 and hence deduce that F(y)=π(1−e−y}{2}  Use this to deduce that for y&amp;gt;0 and a&amp;gt;0  ∫0∞sin⁡xyx(x2+a2)dx=π(1−e−ay)2a2  and  ∫0∞cos⁡xyx2+a2dx=πe−ay2a

Elaine Verrett · Accepted Answer

Since  x1+x2=∫0∞e−ysin⁡(xy)dy,  we have that  I=∫0∞cos⁡bx1+x2dx=∫0∞cos⁡bxxdx∫0∞e−ysin⁡(xy)dy  Changing the order of integration (which can be justified by the standard method) yields  I=∫0∞e−ydy∫0∞sin⁡xyxcos⁡bxdx  The calculation of the integral (a.k.a. the discontinuous Dirichlet factor)  ∫0∞sin⁡xyxcos⁡bxdx={00&amp;lt;y&amp;lt;bπ20&amp;lt;b&amp;lt;y  can be easily reduced to the calculation of the standard Dirichlet integral. Therefore,  I=π2∫b∞e−ydy=π2e−b

Vasquez · Accepted Answer

Another way forward is to use Feynman&#039;s Trick. To that end, let I(a) be the integral I(a,b)=∫0∞e−axcos⁡(bx)1+x2dx (1) for a≥0. Then, for a&amp;gt;0 we have d2I(a,b)db2=−∫0∞e−axx2cos⁡(bx)1+x2dx =I(a,b)−aa2+b2 Therefore, using limb→∞I(a,b)=0 (apply the Riemann-Lebesgue Lemma), we find that for a&amp;gt;0 I(a,b)=C(a)e−|b|+12∫0∞e−|b−x|aa2+x2dx (2) for some function C(a) When b=0 and a→0+, I(a,b) as given by (2) is I(0+,0)=C(0+)+π4, where I(0+,0) as given by (1) is I(0+,0)=π2. Therefore, we find that C(0+)=π4. NSK Finally, when b=1 we have from(1), lima→0+I(a,1)=I(0,1) (3) whereas we have from (2), we have lima→0+I(a,1)=π2e (4) Comparing (3) and (4) yields I(0,1)=π2e as expected!

Suppose that we do not know anything about the complex

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