10. Prove the following results: (a) -k²t-ikx dk e V27 V2t (b) / = F (k) g (k) eikæ dk | f(4) G (y – x) dy, / F (k) g (k) dk = s (1)G(1) dy, (c) 2 a sin x 1 eax (d) sin x * e-a|x| V п (1+а?)> (e) ea"...


10. Prove the following results:<br>(a)<br>-k²t-ikx dk<br>e<br>V27<br>V2t<br>(b) / =<br>F (k) g (k) eikæ dk<br>| f(4) G (y – x) dy,<br>/ F (k) g (k) dk = s (1)G(1) dy,<br>(c)<br>2 a sin x<br>1 eax<br>(d) sin x * e-a|x|<br>V<br>п (1+а?)><br>(e) ea
0 , a V27 (f) exp (-) * exp (-) (-xa). 1 V2a еxp 4b V2(a+b) exp 4(a+b) 4a V2b "/>
Extracted text: 10. Prove the following results: (a) -k²t-ikx dk e V27 V2t (b) / = F (k) g (k) eikæ dk | f(4) G (y – x) dy, / F (k) g (k) dk = s (1)G(1) dy, (c) 2 a sin x 1 eax (d) sin x * e-a|x| V п (1+а?)> (e) ea" * X[0,00) (x) = 1 a > 0 , a V27 (f) exp (-) * exp (-) (-xa). 1 V2a еxp 4b V2(a+b) exp 4(a+b) 4a V2b

Jun 05, 2022
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