Question Number 209923 by OmoloyeMichael last updated on 26/Jul/24
$$\boldsymbol{{Solve}}:\:\int\frac{\boldsymbol{{sin}}\left(\boldsymbol{{x}}!\right)}{\boldsymbol{{x}}!}\boldsymbol{{dx}} \\ $$
Answered by MrGaster last updated on 03/Feb/25
![∫((sin(x!))/(x!))dx=Σ_(k=0) ^∞ ∫_(kπ) ^((k+1)) ((sin(x)!)/(x!)) =Σ_(k=0) ^∞ [(−1)^k ∫_0 ^π ((sin((kπ+t)!))/((kπ+t)!))dt) =Σ_(k=0) ^∞ [(−1)^k (((cos((kπ−t)!))/((kπ+t)!)))∣_0 ^π ] =Σ_(k=0) ^∞ [(−1)^k (((cos((k+1)π!))/((k+1)π!))−((cos(kπ!))/(kπ)))] =lim_(n→∞) [(−1)^n ((cos((n−1)π!))/((n+1)!))+Σ_(k=1) ^n (((−1)^(k=1) )/((k−1)π!))+(1/(π!))] =(1/(π!))+Σ_(k=1) ^∞ (((−1)^(k−1) )/((k−1)π!)) I=∫((sin(x!))/(x!))dx=(1/(π!))+Σ_(k=1) ^∞ (((−1)^(k−1) )/((k−1)π!))](Q216303.png)
$$\int\frac{\mathrm{sin}\left({x}!\right)}{{x}!}\mathrm{d}{x}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{{k}\pi} ^{\left({k}+\mathrm{1}\right)} \frac{\mathrm{sin}\left({x}\right)!}{{x}!} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\left(\left({k}\pi+{t}\right)!\right)}{\left({k}\pi+{t}\right)!}{dt}\right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(−\mathrm{1}\right)^{{k}} \left(\frac{\mathrm{cos}\left(\left({k}\pi−{t}\right)!\right)}{\left({k}\pi+{t}\right)!}\right)\mid_{\mathrm{0}} ^{\pi} \right] \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(−\mathrm{1}\right)^{{k}} \left(\frac{\mathrm{cos}\left(\left({k}+\mathrm{1}\right)\pi!\right)}{\left({k}+\mathrm{1}\right)\pi!}−\frac{\mathrm{cos}\left({k}\pi!\right)}{{k}\pi}\right)\right] \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{cos}\left(\left({n}−\mathrm{1}\right)\pi!\right)}{\left({n}+\mathrm{1}\right)!}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}=\mathrm{1}} }{\left({k}−\mathrm{1}\right)\pi!}+\frac{\mathrm{1}}{\pi!}\right] \\ $$$$=\frac{\mathrm{1}}{\pi!}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\left({k}−\mathrm{1}\right)\pi!} \\ $$$${I}=\int\frac{\mathrm{sin}\left({x}!\right)}{{x}!}\mathrm{d}{x}=\frac{\mathrm{1}}{\pi!}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\left({k}−\mathrm{1}\right)\pi!} \\ $$
Commented by mr W last updated on 23/Mar/25
$${but}\:{the}\:{question}\:{didn}'{t}\:{ask} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({x}!\right)}{{x}!}\mathrm{d}{x} \\ $$