It is possible to prove that ∫0∞eix−e−xxdx=−iπ2

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It is possible to prove that  ∫0∞eix−e−xxdx=−iπ2

twineg4 · Accepted Answer

I think you may simply considerf(α)=∫0+∞e−αx−e−xxdxas a complex variable function with the assumption Re(α)&amp;gt;0. Then:f′(α)=−∫0+∣∞e−αxdx=−1αand f(1)=0, sof(α)=−∫1αdzzSince Re(α)&amp;gt;0, the last complex integral is well defined, and you may define Ref(α) over Ref(α) over {Re(z)≥0}2πiZ by analytic continuation, since Relog⁡α=log||α||. We also have f(α)=f(a―) by the Schwarz&#039; reflection principle andf(α)=−f(1α)by the obvious substitution. Another chance is given by the well-known lemma∫0+∞f(x)dxx=∫0+∞L(f)(s)dsbut we have to be careful with that, since in our case we are considering a Laplace transform on the boundary of its convergence domain.

aquariump9 · Accepted Answer

Note that from Cauchys

Vasquez · Accepted Answer

∫0∞e−ix−e−xxdx=∫0∞(e−ix−e−x)∫0∞e−xtdtdx =∫0∞∫0∞[e−(t+i)x−e−(t+1)x]dxdt=∫0∞(1t+i−1t+1)dt =−ln⁡(i)=−π2i

It is possible to prove that \int_0^{\infty}\frac{e^{ix}-e^{-x}}{x}dx=-i\frac{\pi}{2}

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New Questions in Calculus 1